
JEE Mains 2017 (CodeD)
Test Name: JEE Mains 2017 (CodeD)
Question 1:
It S is the set of distinct values of ‘b’ for which the following system of linear equations
x + y + z = 1
x + ay + z = 1
ax + by + z = 0
has no solution, then S is :
Option 1:  an empty set 
Option 2:  an infinite set 
Option 3:  a finite set containing two or more elements 
Option 4:  a singleton 
Solutions:The given system of equations can be written as
1111a1ab1. Determinant of the above matrix is as follows:
D=1111a1ab1=a12As the system of linear equations has no solution, therefore D = 0.
⇒a12=0⇒a=1For a = 1, the first two planes are coincident to each other. So, x + y + z = 1 and ax + by + z = 0 It is given that S is the set of distinct values of ‘b’ for which the given system of linear equations has no solution. For no solution these two lines are parallel.
⇒1a=1b=11⇒a=1, b=1Therefore, set S has just one value i.e., it’s a singleton set. Hence the correct answer is option (D).
Question 2:
The following statement
(p → q) → [(~ p → q) → q ] is :
Option 1:  a tautology 
Option 2:  equivalent to ~ p → q 
Option 3:  equivalent to p → ~ q 
Option 4:  a fallacy 
Solutions:The truth table for the statement
p→q→~p→q→qis given below:
p  ~p  q  p→q  ~p→q  ~p→q→q  p→q→~p→q→q 
T  F  T  T  T  T  T 
T  F  F  F  T  F  T 
F  T  T  T  T  T  T 
F  T  F  T  F  T  T 
So, the given statement is a tautology.
Hence, the correct answer is option A.
Question 3:
If 5(tan^{2}x – cos^{2}x) = 2cos2x + 9, then the value of cos4x is :
Option 1:  3 5 
Option 2:  13 
Option 3:  29 
Option 4:  79 
Solutions:
5tan2xcos2x=2cos2x+9⇒5sin2xcos2xcos2x=2cos2x+9 ∵tanx=sinxcosx⇒51cos2x1+cos2x1+cos2x2=2cos2x+9 ∵sin2x=1cos2x2 and cos2x=1+cos2x2⇒521cos2x1+cos2x221+cos2x=2cos2x+9 ⇒51cos22x4cos2x=21+cos2x2cos2x+9⇒55cos22x20cos2x=4cos22x+22cos2x+18⇒9cos22x+42cos2x+13=0 ⇒3cos2x+133cos2x+1=0⇒cos2x=13 or cos2x=133Now, cos4x=2cos22x1=79 ∵cos2x≠133Hence, the correct answer is option D.
Question 4:
For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) =
14and P (All the three events occur simultaneously) =
16. Then the probability that at least one of the events occurs, is :
Option 1:  732 
Option 2:  716 
Option 3:  764 
Option 4:  316 
Solutions:It is given that for three events A, B and C, P(Exactly one of A or B occurs) =
14
⇒PA+PB2PA∩B=14 …..(1)
P(Exactly one of B or C occurs) =
14
⇒PB+PC2PB∩C=14 …..(2) P(Exactly one of C or A occurs) =
14
⇒PA+PC2PA∩C=14 …..(3) and P (All the three events occur simultaneously) =
116
⇒PA∩B∩C=16 …..(4) Adding equations 1, 2 and 3, we get
⇒2PA+2PB+2PC2PA∩B2PB∩C2PA∩C=34
⇒PA+PB+PCPA∩BPB∩CPA∩C=38 …(5) The probability that at least one of the events occurs =
PA+PB+PCPA∩BPB∩CPA∩C+PA∩B∩CSo, adding equation (5) and (6), we get
⇒PA+PB+PCPA∩BPB∩CPA∩C+PA∩B∩C=38+116=6+116=716Hence the correct answer is option (B).
Question 5:
Let ω be a complex number such that 2ω + 1 = z where z =
–3. If
1111ω21ω21ω2ω7= 3k, then k is equal to :
Option 1:  –z 
Option 2:  z 
Option 3:  –1 
Option 4:  1 
Solutions:Given:
2ω+1=z where
z=3 and
1111ω21ω21ω2ω7=3kNow
z=3=1×3=i3So,
2ω+1=i3⇒2ω=i31⇒ω=i312 …(1)
We know
1+ω+ω2=0 and ω3=1 …(2)
1111ω21ω21ω2ω7=3k (given)
⇒1111ωω21ω2ω6×ω=3k⇒1111ωω21ω2(ω3)2×ω=3k⇒1111ωω21ω2ω=3k [∵ω3=1]
C1→C1+C2+C3⇒1+1+1111+ω+ω2ωω21+ω+ω2ω2ω=3k ⇒3110ωω20ω2ω=3k from2On expanding we get,
3ω2ω4=3k⇒k=ω2ω4=ω2ωOn putting the value of
ω we get,
1i32–1+i32=i32i32=i3=zHence, the correct answer is option (A).
Question 6:
Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :
Option 1:  2,12 
Option 2:  1,34 
Option 3:  1,34 
Option 4:  2,12 
Solutions:It is given that, the triangle with vertices (k, −3k), (5, k) and (−k, 2) has area 28 sq. units, for any integer k. By area formula, we have
12k3k15k1k21=±28⇒k3k15k1k21=±56⇒kk2+3k5+k+110+k2=±56⇒k22k+15k+3k2+10+k2=±56 ⇒5k2+13k+66 =0 or ⇒5k2+13k46 =0The equation
5k2+13k+66 =0has no solution since
1324×5×66=1691320<0. Now, solve the equation
5k2+13k46 =0. After solving we get the roots of the equation as
k=13±169+92010=13±3310⇒k=2 or 235As k is an integer therefore k = 2. Now, A(2, −6), B(5, 2), C(−2, 2). Equation of altitude dropped from vertex A is x = 2 ….. (i) Equation of altitude dropped from vertex C is 3x + 8y − 10 = 0 …..(ii) Consider the figure below:
Solving (i) and (ii), we get the orthocentre of the triangle as
2,12Hence the correct answer is option (D).
Question 7:
Twenty meters of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flowerbed, is :
Option 1:  12.5 
Option 2:  10 
Option 3:  25 
Option 4:  30 
Solutions:Perimeter of the flower bed in the form of sector = Total length of the wire
⇒2r+l=20⇒2r+rθ=20⇒θ=202rr …1 Area of sector (A)=πr2θ360=12r2·202rr=r10r⇒A=10rr2For maximum or minimum area,
dAdr=0⇒d10rr2dr=0⇒102r=0⇒r=5mPutting the value of r in (1) we get,
θ=20255=20105=2 radian
So, the maximum area comes out to be
A=12×r2×θ=12×52×2=25m2Hence the correct answer is option (C).
Question 8:
The area (in sq. units) of the region {(x, y) : x ≥ 0, x + y ≤ 3, x^{2} ≤ 4y and y ≤ 1 +
x} is :
Option 1:  5912 
Option 2:  32 
Option 3:  73 
Option 4:  52 
Solutions:Given that
x,y:x>0.x+y≤3,x2≤4y and y≤1+xConsider the following figure:
Now,
y≤1+x ⇒y12≤xTherefore, the required area is
A=∫01[1+xx24]dx+∫12[3xx24]dxA=x+23×32x31201+3xx22x31212A=1+23112+62233+12+112A=2+12=52Hence the correct answer is option (D).
Question 9:
If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,
x1=y4=z5is Q, then PQ is equal to :
Option 1:  35 
Option 2:  242 
Option 3:  42 
Option 4:  65 
Solutions:The equation of the line passing through P(1, −2, 3) and parallel to the line given line is
x11=y+24=z35 …..(1) Let R be the point of intersection of the given plane and line (1). Then the coordinates of R are given by (λ + 1, 4λ − 2, 5λ + 3). Now, R(λ + 1, 4λ − 2, 5λ + 3) lies on the plane 2x + 3y − 4z + 22 = 0.
∴2λ+1+34λ245λ+3+22=0⇒6λ+6=0⇒λ=1So, coordinates of R = (1 + 1, 4 × 1 − 2, 5 × 1 + 3) = (2, 2, 8). Now,
PR=212+2+22+832=1+16+25=42So,
PQ=2PR=242Hence, the correct answer is option B.
Question 10:
If for
x∈0,14, the derivative of tan^{–1}
6xx19x3is
x . g(x),then g(x) equals :
Option 1:  91+9×3 
Option 2:  3xx19×3 
Option 3:  3×19×3 
Option 4:  31+9×3 
Solutions:
Let y=tan16xx19×3 for x∈0, 14 y=tan12.3xx13xx2 ∵tan12×1x2=2tan1x =2tan13xx
Now, dydx=2×11+3xx2×3×32×12 =91+9x3x∴ gx=91+9x3Hence, the correct answer is option A.
Question 11:
If (2 + sin x)
dydx+ (y + 1) cos x = 0 and y(0) = 1, then y
π2is equal to :
Option 1:  13 
Option 2:  23 
Option 3:  13 
Option 4:  43 
Solutions:It is given that,
2+sinxdydx+y+1cosx=0.
⇒dydx=y+1cosx2+sinx
⇒dyy+1=cosx2+sinxdxIntegrating both sides,
⇒∫dyy+1=∫cosx2+sinxdx⇒logy+1=log2+sinx+C⇒logy+1+log2+sinx=C⇒logy+12+sinx=C
⇒y+12+sinx=AFor x=0, y=1, ⇒A=4∴y+12+sinx=4For x=π2, ⇒y=13Hence the correct answer is option (A).
Question 12:
Let a vertical tower AB have its end A on the level ground. Let C be the midpoint of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β, then tanβ is equal to
Option 1:  67 
Option 2:  14 
Option 3:  29 
Option 4:  49 
Solutions: Let
∠APC=α and ∠APB=θ.
tanθ=ABAP=x2x=12 tanα=ACAP=x22x=14Let
tanβ=y.
tanθ=tanα+β=tanα+tanβ1tanαtanβ⇒12=14+y114y⇒12=1+4y4y ⇒4y=21+4y⇒4y=2+8y⇒2=9y⇒29=yHence, the correct answer is option (3).
Question 13:
If A =
234 1,then adj (3A^{2} + 12A) is equal to
Option 1:  728463 51 
Option 2:  51638472 
Option 3:  51846372 
Option 4:  726384 51 
Solutions:It is given that,
A=2341. To find 3A^{2 }+ 12A, first find out 3A^{2}.
3A2=323412341=31691213=48273639Now,
12A=24364812Therefore,
3A2 + 12A=48273639+24364812=72638451.
So,
adj3A2+12A=51638472.Hence the correct answer is option (B).
Question 14:
For any three positive real numbers a, b and c, 9(25a^{2} + b^{2}) + 25(c^{2} – 3ac) = 15b(3a + c), Then
Option 1:  b , c and a are in G.P. 
Option 2:  b, c and a are in A.P. 
Option 3:  a, b and c are in A.P. 
Option 4:  a, b and c are in G.P. 
Solutions:We have,
925a2+b2+25c23ac=15b3a+c⇒225a2+9b2+25c245ab15bc75ac=0⇒15a2+3b2+5c215a3b3b5c15a5c=0⇒1215a3b2+3b5c2+5c15a2=0 ⇒15a=3b=5c⇒a=k15, b=k3, c=k5⇒a+b=2c∴b, c, a are in A.P.Hence, the correct answer is option B.
Question 15:
The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines
x11=y+22=z43and
x22=y+11=z+71,is
Option 1:  2074 
Option 2:  1083 
Option 3:  583 
Option 4:  1074 
Solutions:Consider the equation of lines,
x11=y+22=z43 and x22=y+11=z+71. Normal vector is as follows:
i^j^k^123211=5i^+7j^+3k^Since it is given that normal of the plane passing through the point (1, −1, −1) is perpendicular to both of the lines
x11=y+22=z43 and x22=y+11=z+71. Therefore, equation of the plane is
5x1+7y+1+3z+1=0⇒5x+7y+3z+5=0So, the distance of the point (1, 3, −7) from the above plane is as follows:
D=5+2121+525+49+9=1083Hence the correct answer is option (B).
Question 16:
Let I_{n} =
∫tann x dx , (n > 1). If I_{4} + I_{6} = a tan^{5}x + bx^{5} + C, where C is a constant of integration, then the ordered pair (a, b) is equal to
Option 1:  15, 1 
Option 2:  15, 0 
Option 3:  15,1 
Option 4:  15, 0 
Solutions:
In=∫tannx dx=∫tan2x tann2x dx=∫sec2x1tann2x dx=∫tann2x sec2x dx∫tann2x dx=tann1xn1In2 ∫fxnf’xdx=fxn+1n+1+C∴ In=tann1xn1In2 …..1Putting n = 6 in (1), we have
I6=tan5x5I4⇒I6+I4=15tan5x+CComparing this expression with
I6+I4=atan5x+bx5+C, we have
a=15, b=0Thus, the ordered pair (a, b) is
15,0.
Hence, the correct answer is option B.
Question 17:
The eccentricity of an ellipse whose centre is at the origin is
12. If one of its directrices is x = –4, then the equation of the normal to it at
1,32is
Option 1:  2y – x = 2 
Option 2:  4x – 2y = 1 
Option 3:  4x + 2y = 7 
Option 4:  x + 2y = 4 
Solutions: Given one of the directrices is x = −4.
⇒ae=4⇒a=4e⇒a=4×12 Given e=12⇒a=2
Now e2=1b2a2⇒b2=a21e2=a21122=3Equation of ellipse is
x2a2+y2b2=1⇒x24+y23=1Equation of normal at
1,32 is
a2xx1b2yy1=a2b2⇒4×13y32=43⇒4x2y=1Hence, the correct answer is option B.
Question 18:
A hyperbola passes through the point
P 2 , 3and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point :
Option 1:  32, 23 
Option 2:  22, 33 
Option 3:  3, 2 
Option 4:  2, 3 
Solutions:Consider the following figure:
Here, the hyperbola passes through the point P(2, 3) and has foci at (±2, 0). Also, ae = 2. Now, the distance
PF=222+3=942=221Distance of the point P from F’ is
PF’=2+22+3=9+42=22+1
∴PFPF’=2∴a=1 ⇒e=2∵e2=1+b2a2∴b2=3So, equation of hyperbola is
x2y23=1 or 3×2y2=3. Equation of the tangent at
2,3will be
32x3y=3. Therefore, the tangent passes through the point
22,33. Hence, the correct answer is option B.
Question 19:
The function
f : R→12,12defined as
fx=x1+x2, is :
Option 1:  Invertible 
Option 2:  injective but not surjective 
Option 3:  surjective but not injective 
Option 4:  neither injective nor surjective 
Solutions:The function
f : R→12,12is defined as
fx=x1+x2.
fx=x1+x2Diffentiating both sides w.r.t x, we have
f’x=1+x2×1x×2×1+x22=1x21+x22=x1x+11+x22Now, f ‘(x) = 0 for x = 1 and x = −1. f ‘(x) > 0 for −1 < x < 1 and f ‘(x) < 0 for x < −1 and x > 1 Also, f(0) = 0, f(1) =
12and f(−1) =
12
limx→∞fx=limx→∞x1+x2=limx→∞11x+x→∞limx→∞fx=limx→∞x1+x2=limx→∞11x+x→∞So, the graph of the given function is It can be seen that the given function is manyone. So, the given function is not injective. Now, Range of f =
12,12= Codomain of f So, the given function is surjective. Thus, the given function is surjective but not injective.
Hence, the correct answer is option C.
Question 20:
limx→π2cotxcosxπ2x3equals
Option 1:  124 
Option 2:  116 
Option 3:  18 
Option 4:  14 
Solutions:Consider,
limx→π2cotxcosxπ2×3.
limx→π2cotx1sinx8xπ23=limx→π2tanπ2x8π2x1cosπ2xπ2x2=18×1×12=116Hence the correct answer is option (B).
Question 21:
Let
a→=2i^+j^−2k^and
b→=i^+j^. Let
c→ be a vector such that
c→−a→=3,(a→×b→)×c→=3and the angle between
c→and
a→×b→be 30°. Then
a→·c→is equal to
Option 1:  258 
Option 2:  2 
Option 3:  5 
Option 4:  18 
Solutions:Consider
a⇀=2i^+j^2k^, b⇀=i^+j^ and c⇀a⇀=3, a⇀×b⇀×c⇀=3Now,
a⇀×b⇀=2i^2j^+k^⇒ a⇀×b⇀=3We know that,
a⇀×b⇀×c⇀=a⇀×b⇀c⇀sin30°n^⇒ a⇀×b⇀×c⇀=3c⇀×12⇒3=3c⇀×12 Given: a⇀×b⇀×c⇀=3⇒c⇀=2
Now, c⇀a⇀=3Squaring , we get ⇒c⇀2+a⇀22c⇀.a⇀=9⇒4+92a⇀.c⇀=9⇒a⇀.c⇀=2Hence the correct answer is option (B).
Question 22:
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the yaxis passes through the point :
Option 1:  12,12 
Option 2:  12, 12 
Option 3:  12,13 
Option 4:  12, 13 
Solutions:The given curve has the equation y(x – 2)(x – 3) = x + 6. Since the given curve intersects the y axis so, x coordinate will be 0. Therefore, y(0 – 2)(0 – 3) = 0 + 6
⇒y=1So, the point of intersection of the curve with the y axis is (0, 1). Now slope of the curve y(x – 2)(x – 3) = x + 6 can be calculated as
⇒y=x+6(x – 2)(x – 3)⇒dydx=x25x+6x+62x5×25x+62At (0, 1) the slope will be
dydx=6–3036⇒dydx=1Thus, slope of the curve is 1. So, the slope of the normal to this curve will be −1. So, the equation of the normal is
yy1=mxx1
⇒y1=1x0⇒y1=x⇒x+y1=0Thus,
12,12 satisfies the given equation of the normal. Hence, the correct answer is option B.
Question 23:
If two different numbers are taken from the set {0, 1, 2, 3,….., 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is
Option 1:  655 
Option 2:  1255 
Option 3:  1445 
Option 4:  755 
Solutions:Let
A=0,1,2,…,10n(s) =
C211. Let B be the number of selecting two numbers whose sum as well as absolute difference both are multiple of 4. If a and b are two numbers selected from the given set then,
Sum=a+bAbsolute difference=abSo, B = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)} n(B) = 6
∵ the order does not matters and which is clearly indicated in the sample space
i.e 11C2 Thus, P=nBnS=6C211=655. Hence, the correct answer is option A.
Question 24:
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is
Option 1:  485 
Option 2:  468 
Option 3:  469 
Option 4:  484 
Solutions:According the question given :
Man X’ friends : 4L and 3M Wife Y’s friends : 3L and 4M
When a party is thrown there should be 3M and 3L on the whole. But 3 friends each of X and Y should appear.
Therefore, the no.of possibilities of throwing a party
X 3M 1L,2M 1M,2L 3L
Y 3L 2L,1M 2M,1L 3M
^{3}C_{3}* ^{3}C_{3} + ^{4}C_{1}* ^{3}C_{2} * ^{4}C_{1}* ^{3}C_{2} + ^{3}C_{1}* ^{4}C_{2} * ^{4}C_{2}* ^{3}C_{1 }+ ^{4}C_{3}* ^{4}C_{3} = 485
Hence , the correct answer is option A.
Question 25:
The value of (^{21}C_{1} – ^{10}C_{1}) + (^{21}C_{2} – ^{10}C_{2}) + (^{21}C_{3} – ^{10}C_{3}) + (^{21}C_{4} – ^{10}C_{4}) +……..+ (^{21}C_{10} – ^{10}C_{10}) is
Option 1:  2^{21} – 2^{11} 
Option 2:  2^{21} – 2^{10} 
Option 3:  2^{20} – 2^{9} 
Option 4:  2^{20} – 2^{10} 
Solutions:S = (^{21}C_{1} – ^{10}C_{1}) + (^{21}C_{2} – ^{10}C_{2}) + (^{21}C_{3} – ^{10}C_{3})+(^{21}C_{4} – ^{10}C_{4})+(^{21}C_{5} – ^{10}C_{5})+(^{21}C_{6} – ^{10}C_{6})+(^{21}C_{7} – ^{10}C_{7})+(^{21}C_{8} – ^{10}C_{8})+(^{21}C_{9} – ^{10}C_{9})+(^{21}C_{10} – ^{10}C_{10})
S = ( ^{21}C_{1} + ^{21}C_{2} + ^{21}C_{3 }+_{ }^{21}C_{4} + ^{21}C_{5 }+ ^{21}C_{6 }+ ^{21}C_{7 }+ ^{21}C_{8 }+ ^{21}C_{9 }+ ^{21}C_{10} ) – ( ^{10}C_{1 }+ ^{10}C_{2 }+ ^{10}C_{3 }+_{ }^{10}C_{4 }+_{ }^{10}C_{5 }+^{10}C_{6 }+_{ }^{10}C_{7 }+^{10}C_{8 }+ ^{10}C_{9 }+^{10}C_{10 })
S_{1 }= ( ^{21}C_{1} + ^{21}C_{2} + ^{21}C_{3 }+_{ }^{21}C_{4} + ^{21}C_{5 }+ ^{21}C_{6 }+ ^{21}C_{7 }+ ^{21}C_{8 }+ ^{21}C_{9 }+ ^{21}C_{10} )
=
12( ^{21}C_{1} + ^{21}C_{2} +…..+ ^{21}C_{20 })
=
12( ^{21}C_{0} + ^{21}C_{1} + ^{21}C_{2} +…..+ ^{21}C_{20 }+ ^{21}C_{21} – 2 ) {
∵ ^{n}C_{0}+nC_{1}+^{n}C_{2} +…..+^{n}C_{n} =
2n}
=
12(
2212 ) =
220 – 1S_{2 }=_{ }( ^{10}C_{1 }+ ^{10}C_{2 }+ ^{10}C_{3 }+_{ }^{10}C_{4 }+_{ }^{10}C_{5 }+^{10}C_{6 }+_{ }^{10}C_{7 }+^{10}C_{8 }+ ^{10}C_{9 }+^{10}C_{10 })
= ( ^{10}C_{0 }+^{10}C_{1 }+ ^{10}C_{2 }+…..+ ^{10}C_{10 }– 1 )
=
210 – 1
∴ S_{1 }–_{ }S_{2 }=_{ }
220 – 210Hence , the correct answer is option D
Question 26:
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, onebyone, with replacement, then the variance of the number of green balls drawn is
Option 1:  125 
Option 2:  6 
Option 3:  4 
Option 4:  625 
Solutions:Total = 25 balls (15 green and 10 yellow) Let, p be the probability of drawing green ball and q be the probability of drawing yellow ball (with replacement)
p=1525=35
q=1025=25A random variable X which takes values 0,1,2….10 ( 10 balls to be drawn) is said to follow binomial distribution with parameters n and p and it is also satisfying the condition p+q = 1 and in that case Variance = npq. n = 10 ,
p=35, q=25Variance =
10×35×25=125Hence, the correct answer is option A.
Question 27:
Let a, b, c ∈ R. If f(x) = ax^{2} + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy,∀x, y ∈ R, then
∑10n=1fnis equal to
Option 1:  330 
Option 2:  165 
Option 3:  190 
Option 4:  225 
Solutions: Let a,b,c
∈R. If f(x) = ax^{2} + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy,
∀ x,y
∈ R, then
∑n=110f(n) is equal to
f(x) = ax^{2} + bx + c f(x + y) = f(x) + f(y) + xy a(x + y)^{2} + b(x+ y) + c = ax^{2} + bx + c + ay^{2} + by + c + xy 2axy = c + xy
∀ x, y
∈R (2a – 1)xy – c = 0
∀ x, y
∈R
⇒c=0
∵ independent term is zero both sides
⇒a=12
∵ coefficient of quadratic term is zero both sides and thereby equating (2a – 1)=0
a + b + c = 3
12 + b + 0 = 3 b=
52 ∴f(x)=12×2+52x∑n=110f(n)=12∑n=110n2+52∑n=110n ⇒12x10x11x216+52x10x112=330Hence , the correct answer is option A
Question 28:
The radius of a circle, having minimum area, which touches the curve y = 4 – x^{2} and the lines, y = x is
Option 1:  22+1 
Option 2:  221 
Option 3:  421 
Option 4:  42+1 
Solutions:The radius of a circle, having minimum area, which touches the curve y = 4 – x^{2} and the lines, y = x is
The equation of parabola ^{ }y = 4 – x^{2 }on rearranging in standard form x^{2 }= – (y – 4) and comparing it with x^{2} = – 4ay The parabola faces downwards due to negative sign and since
a=14 and vertex will be at
y=4
let radius of circle be r, its center lies on yaxis as yaxis bisects the 2 rays of y = x due to symmetry center of the circle must be on yaxis let center be (0, k) Length of perpendicular from (0, k) to y = x,
r=k2
∴Equation of circle
x2+(yk)2=k22solving circle and parabola
4y+y22ky+k22=0y2(2k+1)y+k22+4=0Because circle just touches the parabola , so the quadratic equation formed by combining the quadratic equations of parabola and circle will give us the common solution (point where they meet)
∴D = 0
(2k+1)2=4k22+44k2+4k+1=2k2+16On solving we get
k=4+1364Therefore radius =
k2radius
r =4+13642 r≈1.354Option 1
2(2+1)≈4.828 Option 2
2(21)≈0.8284Option 3
4(21)≈1.656Option 4
4(2+1)≈9.656None of the option is exactly correct , however option C is the closest value to
r≈1.354Hence , the correct answer is option C.
Question 29:
If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1)(x + 2) +…..+ (x +
n–1)(x + n) = 10n has two consecutive integral solutions, then n is equal to
Option 1:  12 
Option 2:  9 
Option 3:  10 
Option 4:  11 
Solutions:Ans
Given equation,
x(x+1)+(x+1)(x+2)+(x+2)(x+3)+…+{x+(n1)}{x+n}=10n
first term, x(x+1)= x2 + x + 0Second term, (x+1)(x+2) = x2 +3x + 2Third term, (x+2)(x+3) = x2 +5x + 6 Fourth term, (x+3)(x+4) = x2 +7x + 12……… ……………nth term, {x+(n1)}{x+n}=x2 +(2n1)x+n(n1)Adding all the terms,
x(x+1)+(x+1)(x+2)+…+{x+(n1)}{x+n}=nx2+(1+3+5+…to n terms)x+(0+2+6+12+…to n terms) =
nx2+n2x+∑r=1nn(n1) =
nx2+n2x+n(n+1)(2n+1)6n(n+1)2 =
nx2+n2x+n(n21)3Therefore, the given equation can be written as,
nx2+n2x+n(n21)3=10n
⇒x2+nx+n21310=0⇒x2 +nx + n2313=0Let the roots are
α and
β then
α+β=n and αβ=1 (given)
(α+β)2=(αβ)2+4αβ⇒n2=12+4.n2313
⇒3n2=3+4n2124⇒n2=121⇒n=11.Hence, the correct answer is option D.
Question 30:
The integral
∫π43π4dx1+cosxis equal to
Option 1:  −2 
Option 2:  2 
Option 3:  4 
Option 4:  −1 
Solutions:The integral
∫π43π4dx1+cos x is equal to
I= ∫π43π4dx1+cos x …(1)Using property
∫abf(x) dx=∫abf(a+bx) dx
I= ∫π43π4dx1cos x …(2)Adding (1) and (2)
2I= ∫π43π42dx1cos2x ⇒I= ∫π43π4dxsin2x⇒I= ∫π43π4cosec2x dxI= cot x π43π4 = 2Hence, the correct answer is option B
Question 31:
A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :
Option 1:  t=Tlog1.3 
Option 2:  t=T2log2log1.3 
Option 3:  t=Tlog 1.3log 2 
Option 4:  t=T log 1.3 
Solutions:Let N_{A }be the total number of A present initially. Then after decaying into B, at time t, the number of A becomes equal to
NA = NoeλtTherefore, number of B at time t is
NB = No(1 – eλt )Now,
NBNA = No(1eλt)NoeλtNBNA = eλt – 1Given, NBNA = 0.3Therefore, 0.3 = eλt – 11.3 = eλtλt = ln(1.3) …(i)Now, it is given that half life of radioactive nucleus A is T which is equal toT = ln(2)λ⇒λ = ln(2)T … (ii)Putting (ii) in (i)t = ln(1.3) ln(2)THence, the correct answer is option C.
Question 32:
The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25 × 10^{−2} m rise of water, h = 1.45 × 10^{−2} m.
Using g = 9.80 m/s^{2} and the simplified relation
T=rhg2×103N/m, the possible error in surface tension is closest to:
Option 1:  10% 
Option 2:  0.15% 
Option 3:  1.5% 
Option 4:  2.4% 
Solutions:Here, the information of least count of D and h measurement are not given. So, we will use maximum.
Permissible error in D and h = place value of last digit.
D = 1.25 × 10^{−2} m
So,
∆D = 0.01 × 10−2 mand, â€‹h = 1.45 × 10^{−2} m
So,
∆h = 0.01 × 10−2 m T =rgh2 × 103∆TT = ∆rr +∆hh = ∆DD +∆hh∆TT = 0.01 × 1021.25 × 102 +0.01 × 1021.45 × 102∆TT = (1125 +1145 ) × 100 %∆TT = ( 0.008 + 0.0069) × 100 %∆TT = 1.49 % ≈ 1.5 %Hence, the correct answer is option C.
Question 33:
An electron bean is acceleration by a potential difference V to hit a metallic target to produce Xrays. It produces continuous as well as characteristic Xrays. If λ_{min} is the smallest possible wavelength of Xray in the spectrum, the variation of log λ_{min} with log V is correctly represented in:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Energy of electron beam can be given as
eV = hcλ⇒ λmin = 12400eVTaking log on both sides⇒ log λmin = log (12400) – log (e) – log (V)log λmin = C – log (V)which is similar to line equationY = C – mx Hence, the correct answer is option B.
Question 34:
The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum?
Option 1:  32 
Option 2:  32 
Option 3:  32 
Option 4:  1 
Solutions:The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is given to be I which is equal to
I = Ml212 + MR24 …(i)where, M is the mass of the cylinder.
Now, we knowMass of the body = volume of the body × density of the body⇒M =(π R2l) ρwhere, ρ = density of the bodyor, R2= Mπ lρ …(ii)Putting (ii) in (i)I = Ml212 + M24π lρ …(iii)Now, differentiating (iii) w.r.t ldIdl= M12 (2l) – M24π l2ρ …(iv)For moment of inertia to be minimum, (iv) should be equal to 0.Hence, M12 (2l) – M24π l2ρ = 0l6 = M4π l2ρl3 = 3M2π ρ = 3(π R2l) ρ2π ρ … using (ii)l2 = 3R22l2 R2 = 32l R = 32Hence, the correct answer is option B.
Question 35:
A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. the angular acceleration of the rod when it makes an angle θ with the vertical is
Option 1:  2g3lcos θ 
Option 2:  3g2lsin θ 
Option 3:  2g3lsin θ 
Option 4:  3g2lcos θ 
Solutions: Taking torque about the pivot
τ = I αMgsinθl2 = Ml23α α = 3g2lsinθHence, the correct answer is option B.
Question 36:
C_{p} and C_{v} are specific heats at constant pressure and constant volume respectively. It is observed that C_{p} − C_{v} = a for hydrogen gas C_{p} − C_{v} = b for nitrogen gas The correct relation between a and b is:
Option 1:  a = 28 b 
Option 2:  a=114b 
Option 3:  a = b 
Option 4:  a = 14 b 
Solutions:We know that
C_{ }= M_{o}_{ }S
For H_{2} as well as N_{2}
C_{p} – C_{v}_{ }= R where, C_{p} and C_{v} are molar specific heat capacities.
M_{o}_{ }S_{p} – M_{o}_{ }S_{v}_{ }= R
Sp – Sv = R Mo
For H_{2} gas
Sp – Sv = R 2 = aand for N_{2}_{ }gas
Sp – Sv = R 28 = bSo,
a b = R2R28 = 14Therefore,
a = 14bHence, the correct answer is option D.
Question 37:
A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75° C. T is given by: (Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm° C)
Option 1:  825° C 
Option 2:  800° C 
Option 3:  885° C 
Option 4:  1250° C 
Solutions:Heat given by the copper ball = heat taken by the water + heat taken by the calorimeter system
Therefore, (100) (0.1) (T – 75) = (170) (1) (75 – 30) + (100) (0.1) (75 – 30)
10T – 750 = 8100
10T = 8850
T = 885°C
Hence, the correct answer is option C.
Question 38:
In amplitude modulation, sinusoidal carrier frequency used is denoted by ω_{c} and the signal frequency is denoted by ω_{m}. The bandwidth (âˆ†ω_{m}) of the signal is such that âˆ†ω_{m}<< ω_{c}. Which of the following frequency is not contained in the modulated wave?
Option 1:  ω_{c} − ω_{m} 
Option 2:  ω_{m} 
Option 3:  ω_{c} 
Option 4:  ω_{m} + ω_{c} 
Solutions:The amplitude modulated signal, c_{m}(t) is given as
cm (t) = Ac sin ωc t + μAc2cos (ωc – ωm ) t – μAc2cos (ωc + ωm ) tSo, it can be inferred from above equation that frequency of ω_{m }is not contained in the modulated wave.
Hence, the correct answer is option B.
Question 39:
The temperature of an open room of volume 30 m^{3} increased from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 10^{5} Pa. If n_{i} and n_{f} are the number of molecules in the room before and after heating, then n_{f} − n_{i} will be :
Option 1:  −2.5 × 10^{25} 
Option 2:  −1.61 × 10^{23} 
Option 3:  1.38 × 10^{23} 
Option 4:  2.5 × 10^{25} 
Solutions:Using ideal gas equation,
PV= NRT where n is the number of moles.
Before heating, the given temperature is T_{i}_{ }= 273 + 17 = 290 K
PoVo = NiR × 290 …(i)
After heating, the given temperature is T_{i}_{ }= 273 + 27 = 300 K
PoVo = NfR × 300 …(ii)
From equation (i) and (ii)
Nf – Ni = PoVoR×300 – PoVoR×290Therefore, difference in number of moles
Nf – Ni = – 10 PoVoR×290×300Hence,
nf – ni = – PoVo × 10R×290×300 × 6.023× 1023putting Po = 10^{5} PA and V_{o} = 30 m^{3}
Therefore, number of molecules n_{f} – n_{i} is
n_{f} – n_{i} = – 2.5 × 10^{25}
Hence, the correct answer is option A.
Question 40:
In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:
Option 1:  15.6 mm 
Option 2:  1.56 mm 
Option 3:  7.8 mm 
Option 4:  9.75 mm 
Solutions:Fringe width is given as
β1 = λDd where, d is the slits’ seperation and D is the separation between the screen and the slitβ1= 650 × 109 × 1.55 × 104 = 650 × 105 × 0.3 = 1.95 mmAlso,β2 = 520 × 109 × 1.55 × 104 = 520 × 105 × 0.3 = 1.56 mm4β1 = 5β2 = 7.8 mmHence, the correct answer is option C.
Question 41:
A particle A of mass m and initial velocity υ collides with a particle B of mass
m2which is at rest. The collision is head on, and elastic. The ratio of the deBroglie wavelengths λ_{A} to λ_{B} after the collision is:
Option 1:  λAλB=12 
Option 2:  λAλB=13 
Option 3:  λAλB=2 
Option 4:  λAλB=23 
Solutions: By conservation of linear momentum,
mv = mvA + m2vB …(i)and by law of collisione = vB – vAv = 1 …(ii) From (i) and (ii)2(vB – vA) = 2vA + vB2vB – vB = 4vAvB= 4vAλAλB = mBvBmAvA =m2m × 4 = 2Hence, the correct answer is option C.
Question 42:
A magnetic needle of magnetic moment 6.7 × 10^{−2} Am^{2} and moment of inertia 7.5 × 10^{−6} kg m^{2} is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:
Option 1:  8.76 S 
Option 2:  6.65 S 
Option 3:  8.89 S 
Option 4:  6.98 S 
Solutions:Given, magnetic moment, M = 6.7 × 10^{−2} A m^{2}
Moment of inertia, I = 7.5 × 10^{−6} kg m^{2}
and magnetic field, B = 0.01 T
Since, time period of oscillation is given as
T = 2πIMB = 2π7.5 × 1066.7 × 10 2 × 10 2 = 2π × 10175 67To complete 10 oscillations, time taken, t , will bet = 10 Tt = 10 × 2π × 10175 67 = 6.65 secHence, the correct answer is option B.
Question 43:
An electric dipole has a fixed dipole moment
→p, which makes angle θ with respect to xaxis. When subjected to an electric field
E→1=Ei^, it experiences a torque
T→1=τk^. When subjected to another electric field
E→2=3E1j^it experiences a t orque
T→2=T→1. The angle θ is:
Option 1:  90° 
Option 2:  30° 
Option 3:  45° 
Option 4:  60° 
Solutions:
We know torque is given as,
τ→ = p→ × E→
Since, both the torque in the given problem is related asT2→ = T1→thus, T1→ + T2 →= 0τk^ + ( τk^) = 0(pxi^ + pyj^) × (Ei ^+ 3Ej^) = 0px × 3Ek ^+ py × E(k ^)= 0Ek ^(3px – py) = 0 pypx = 3Therefore, tanθ = 3θ = 60oHence, the correct answer is option D.
Question 44:
In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is:
Option 1:  275 Wb 
Option 2:  200 Wb 
Option 3:  225 Wb 
Option 4:  250 Wb 
Solutions:As we know emf induced (considering magnitude) in a coil due to change in magnetic flux is given as
e = dϕdt …(i)
Therefore, current induced in the coil of resistance R is
I = eRI =1R dϕdt⇒dϕ = RIdtor, ϕ = R∫Idtor, ϕ = R × charge induced in the coil or, ϕ = R × area of given currenttime graph …(i)Now, area of given currenttime graph = 12 × 10 × 0.5 = 2.5 coloumb …(ii)Putting R = 100 Ω and (ii) in (i)ϕ = 100 × 2.5 = 250 WbHence, the correct answer is option D.
Question 45:
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be:
Option 1:  18 J 
Option 2:  4.5 J 
Option 3:  22 J 
Option 4:  9 J 
Solutions: The acceleration, a, produced in the particle of mass 1 kg is
a=Fm=6t1Since, a =dvdt⇒dvdt=6tdv=6tdt∫0vdv=6∫0ttdtv=6t2201=3Therefore, the work done by the force during the first 1 sec. isW=∆KE=KfKi⇒W=12(1)(3)2=4.5 JHence, the correct answer is option B.
Question 46:
Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ_{1}/λ_{2}, is given by:
Option 1:  r=13 
Option 2:  r=43 
Option 3:  r=23 
Option 4:  r=43 
Solutions:Using,
∆E = hcλfor
λ1
(E)(2E)=hcλ1 …(i)for λ2(E)(43E)=hcλ2 …(ii)Equation (ii)/(i)43121=λ1λ2λ1λ2=13Hence, the correct answer is option A.
Question 47:
In the above circuit the current in each resistance is:
Option 1:  0 A 
Option 2:  1 A 
Option 3:  0.25 A 
Option 4:  0.5 A 
Question 48:
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions: Here, acceleration on body is constant.
a=dvdt=9.8Therefore, the slope of velocity time graph is also a negative constant.
Hence, the correct answer is option D.
Question 49:
A capacitance of 2μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:
Option 1:  32 
Option 2:  2 
Option 3:  16 
Option 4:  24 
Solutions:To hold 1 kV potential difference, minimum four capacitors are required in series. Therefore, capacitance for a series =
14
μF So, for equivalent capacitance to be 2
μF, 8 parallel combinations are required.
Hence, the minimum number of capacitors required is 32 (
8 × 4 ).
Hence, the correct answer is option A.
Question 50:
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:
Option 1:  CEr1r1+r 
Option 2:  CE 
Option 3:  CEr1r2+r 
Option 4:  CEr2r+r2 
Solutions: At steady state, the capacitor will behave as an open circuit. So the current through r_{1} will be zero and hence the branch containing r_{1} can be ignored while considering the equivalent circuit.
Current in the circuit,
I=Er+r2Potential difference across r_{2} , V= Ir_{2}
⇒V=Er2r+r2
Charge on capacitor, Q=CV
Q=CEr2r+r2Hence, the correct answer is option D.
Question 51:
In a common emitter amplifier circuit using an npn transistor, the phase difference between the input and the output voltages will be:
Option 1:  180° 
Option 2:  45° 
Option 3:  90° 
Option 4:  135° 
Solutions:
∆Vcc=∆VCE +RL∆IC=0or ∆VCE=RL∆ICThe change in VCE is the output voltage vo. vo=∆VCE=β RL∆IBThe voltage gain of the amplifier is Av=v0vi=∆VCEr∆IB=βacRLrThe negative voltage shows that the output voltage is opposite in phase with the input voltage.Hence, the correct answer is option A.
Question 52:
Which of the following statements is false ?
Option 1:  Krichhoff’s second law represents energy conservation. 
Option 2:  Wheatstone bridge is the most sensitive when all the four resistance are of the same order of magnitude. 
Option 3:  In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed. 
Option 4:  A rheostat can be used as a potential divider. 
Solutions:In a balanced wheat stone bridge if the position of cell and galvanometer is exchanged the null point remains same.
Hence, the correct answer is option C.
Question 53:
A particle is executing simple harmonic motion with a time period T. At time = 0, it is at its position of equilibrium. The kinetic energytime graph of the particle will look like:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
x=Asin (ωt+φ)ϕ=0,πx=±Asin(ωt)KE=12mω2(A2x2)KE=12KA2cos2 (ωt) Hence, the correct answer is option A.
Question 54:
An observer is moving with half the speed of light towards stationary microwave source emitting waves at frequency 10GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 10^{8} ms^{−1})
Option 1:  15.3 GHz 
Option 2:  10.1 GHz 
Option 3:  12.1 GHz 
Option 4:  17.3 GHz 
Solutions:
ν’=ν1+vc1vcν’=ν1+12112=3νν’=10×1.73=17.3 GHzHence, the correct answer is option D.
Question 55:
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by factor of:
Option 1:  181 
Option 2:  9 
Option 3:  19 
Option 4:  81 
Solutions:Volume of man becomes (9)^{3 }times. Since density is same, so weight of man becomes (9)^{3 }times. Cross sectional area of leg becomes (9)^{2 }times
Stress=WeightArea=9392=9 timesHence, the correct answer is option B.
Question 56:
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale defection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 − 10 V is:
Option 1:  4.005 ×10^{3} Ω 
Option 2:  1.985 × 10^{3} Ω 
Option 3:  2.045 × 10^{3} Ω 
Option 4:  2.535 × 10^{3} Ω 
Solutions:Full scale current, I_{g}_{ }= 5 mA Resistance of galvanometer, G = 15 Ω
R = VIgG= 105×10315= 200015= 1.985 × 103 ΩHence, the correct answer is option B.
Question 57:
The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth’s radius)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
g=GMdR3 d<Rg=GMR2 d⩾R Hence, the correct answer is option A.
Question 58:
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:
Option 1:  3PKα 
Option 2:  P3αK 
Option 3:  PαK 
Option 4:  3αPK 
Solutions:
∆VV=γ∆θ=3α∆θ∆p=K∆VVp=K(3α∆θ)∆θ=p3αKHence, the correct answer is option B.
Question 59:
A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is:
Option 1:  real and at a distance of 6 cm from the convergent lens 
Option 2:  real and at a distance of 40 cm from convergent lens 
Option 3:  virtual and at a distance of 40 cm from convergent lens 
Option 4:  real and at distance of 40 cm from the divergent lens 
Solutions:Image formed by first lens is I_{1} which is 25 cm left of diverging lens. For second lens u = 40 cm (i.e. at 2F), so final image will be 40 cm right of converging lens. Hence image will be real.
Hence, the correct answer is option B.
Question 60:
A body of mass m = 10^{−2} kg is moving in a medium and experiences a frictional force F = −kv^{2}. Its initial speed is v_{0} = 10 ms^{−1}. If after 10 s, its energy is
18mv02, the value of k will be:
Option 1:  10^{−1} kg m^{−1} s^{−1} 
Option 2:  10^{−3} kg m^{−1} 
Option 3:  10^{−3} kg s^{−1} 
Option 4:  10^{−4} kg m^{−1} 
Solutions:
F = kv2mdvdt = kv2∫vovv2dv =∫0tkmdtAlso, after 10 s,KE = 12mv2 = 18mvo2⇒ v= vo2Therefore, 1vvovo/2=kmt2vo1vo=kmt⇒k=mvot=10210×10=104 kg m1Hence, the correct answer is option D.
Question 61:
1 gram of a carbonate (M_{2}CO_{3}) on treatment with excess HCl produces 0.01186 mole of CO_{2}. The molar mass of M_{2}CO_{3} in g mol^{–1 }is :
Option 1:  84.3 
Option 2:  118.6 
Option 3:  11.86 
Option 4:  1186 
Solutions:Given: Mass of carbonate = 1 g Moles of carbon dioxide produced on reaction with hydrogen chloride = 0.01186 mol
To find: Molar mass of the carbonate = ?
Suppose, Molar mass of the metal carbonate (M_{2}CO_{3}) = M g/mol
The balanced chemical reaction between the carbonate and hydrogen chloride can be written as: M_{2}CO_{3} + 2 HCl → 2 MCl_{2 }+ H_{2}O + CO_{2}
According to this reaction,
Moles of M_{2}CO_{3} needed to produce 1 mole of CO_{2} = 1 mol Moles of M_{2}CO_{3} needed to produce 0.01186 moles of CO_{2} = 0.01186 mol
Molar Mass of M_{2}CO_{3} =
Mass of M2CO3Moles of M2CO3=1 g0.01186 mol = 84.3 g/molHence, the correct answer is option (1).
Question 62:
Given : C_{(graphite)} + O_{2}(g) → CO_{2}(g) ; Δ_{r}H° = – 393.5 kJ mol^{–1 }; H_{2}(g) +
12O_{2}(g) → H_{2}O(l) ; Δ_{r}H° = –285.8 kJ mol^{–1 }; CO_{2}(g) + 2H_{2}O(l) → CH_{4}(g) + 2O_{2}(g) ; Δ_{r}H° = + 890.3 kJ mol^{–1} Based on the above thermochemical equations, the value of Δ_{r}H° at 298 K for the reaction C_{(graphite)} + 2H_{2}(g) → CH_{4}(g) will be :
Option 1:  +144.0 kJ mol^{–1} 
Option 2:  –74.8 kJ mol^{–1} 
Option 3:  –144.0 kJ mol^{–1} 
Option 4:  +74.8 kJ mol^{–1} 
Solutions:Given: C(graphite) + O_{2}(g) → CO_{2}(g) ; Δ_{r}Hº = – 393.5 kJ mol^{–1} …(I) H_{2}(g) + 1/2 O_{2}(g) → H_{2}O(l) ; Δ_{r}Hº = – 285.8 kJ mol^{–1} …(II) CO_{2}(g) + 2 H_{2}O(l) → CH_{4}(g) + 2 O_{2}(g) ; Δ_{r}Hº = + 890.3 kJ mol^{–1} …(III)
To Find: Δ_{r}Hº at 298 K for the reaction C(graphite) + 2 H_{2}(g) → CH_{4}(g)
The above reaction can be obtained as: Δ_{r}Hº = I + (2×II) + III Δ_{r}Hº = (− 393.5) + (− 285.8 × 2) + 890.3 Δ_{r}Hº = 890.3 − (393.5 + 571.6) Δ_{r}Hº = 890.3 − 965.1 Δ_{r}Hº = − 74.8 kJ / mol
Hence, the correct answer is option (2).
Question 63:
The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (K_{f }for benzene = 5.12 K kg mol^{–1})
Option 1:  80.4% 
Option 2:  74.6% 
Option 3:  94.6% 
Option 4:  64.6% 
Solutions:Given: Depression in freezing point (ΔT_{f}) = 0.45 ºC Mass of acetic acid (CH_{3}COOH) = 0.2 g Molar mass of acetic acid (CH_{3}COOH) = 60 g/mol Mass of benzene (C_{6}H_{6}) = 20 g Molal depression constant (K_{f}) = 5.12 K kg/mol Number of molecules of acetic acid that associate in benzene = 2
To find: Degree of association for acetic acid (α) = ?
The relation between degree of association and van’t Hoff factor is given as:
α = i11n1van’t Hoff factor can be obtained from the expression for depression in freezing point:
i = ΔTf × MCH3COOH × mC6H6Kf × mCH3COOH × 1000i = 0.45 × 60 × 205.12 × 0.2 × 1000 = 0.53Substituting this value of i in the expression for degree of association we get:
α = 0.531121 = 0.94Percentage association of acetic acid in benzene = 94%
Hence, the correct answer is option (3).
Question 64:
The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%) ; and Nitrogen (2.6%). The weight which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is :
Option 1:  37.5 kg 
Option 2:  7.5 kg 
Option 3:  10 kg 
Option 4:  15 kg 
Solutions:75 kg person contain 10% hydrogen. Mass of hydrogen =
=10100×75 = 7.5 kgIf all ^{1}H atom are replaced by ^{2}H, the mass of hydrogen become twice. So, replacing ^{1}H by ^{2}H would replace 7.5 kg with 15 kg. Therefore, net gain is 7.5 kg.
Hence, the correct answer is option (2).
Question 65:
ΔU equal to :
Option 1:  Isobaric work 
Option 2:  Adiabatic work 
Option 3:  Isothermal work 
Option 4:  Isochoric work 
Solutions:Using first law of thermodynamics,
âˆ†U = q + w
For adiabatic process, q = 0
i.e. âˆ†U = w
Therefore, âˆ†U is equal to adiabatic work.
Hence, the correct answer is option (2).
Question 66:
The formation of which of the following polymers involves hydrolysis reaction?
Option 1:  Bakelite 
Option 2:  Nylon 6,6 
Option 3:  Terylene 
Option 4:  Nylon 6 
Solutions:Nylon 6 is prepared by hydrolysis of caprolactum. The reaction can be shown as:
Hence, the correct answer is option (4).
Question 67:
Given
ECl2/Cl°=1.36 V, ECr3+/Cr°=0.74 VECr2O72/Cr3+ °=1.33 V, EMnO4/Mn2+°=1.51 VAmong the following, the strongest reducing agent is :
Option 1:  Mn^{2+ } 
Option 2:  Cr^{3+} 
Option 3:  Cl^{–} 
Option 4:  Cr 
Solutions:The strongest reducing agent is determined by the least value of reduction potential. Therefore,
ECr3+/Cr°=0.74 Vis the strongest reducing agent.
Hence, the correct answer is option (4).
Question 68:
The Tyndall effect is observed only when following conditions are satisfied :
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particles is not much smaller than the wavelength of the light used
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
Option 1:  (b) and (d) 
Option 2:  (a) and (c) 
Option 3:  (b) and (c) 
Option 4:  (a) and (d) 
Solutions:Tyndall effect is observed only when the diameter of the dispersed particles is not much smaller than the wavelength of the light used and the refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.
Hence, the correct answer is option (1).
Question 69:
In the following reactions, ZnO is respectively acting as a/an :
(a) ZnO + Na_{2}O → Na_{2}ZnO_{2}
(b) ZnO + CO_{2} → ZnCO_{3}
Option 1:  base and base 
Option 2:  acid and acid 
Option 3:  acid and base 
Option 4:  base and acid 
Solutions:(a) ZnO + Na_{2}O → Na_{2}ZnO_{2}
(b) ZnO + CO_{2} → ZnCO_{3}
In equation (a), ZnO behaves like acid while in equation (b), ZnO behaves like base.
Hence, the correct answer is option (3).
Question 70:
Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Reducing sugars are those in which there is presence of free anomeric –OH group.
Hence, the correct answer is option (3)
Question 71:
The major product obtained in the following reaction is :
Option 1:  C_{6}H_{5}CH = CHC_{6}H_{5} 
Option 2:  (+) C_{6}H_{5}CH(O^{t}Bu)CH_{2}C_{6}H_{5} 
Option 3:  (–) C_{6}H_{5}CH(O^{t}Bu)CH_{2}C_{6}H_{5} 
Option 4:  (±) C_{6}H_{5}CH(O^{t}Bu)CH_{2}C_{6}H_{5} 
Solutions:
Hence, the correct answer is option (1).
Question 72:
Which of the following species is not paramagnetic?
Option 1:  CO 
Option 2:  O_{2} 
Option 3:  B_{2} 
Option 4:  NO 
Solutions:Molecular Species
Number of Electrons  MOT Electronic Configuration  Magnetic Nature  
CO  6 + 8 = 14  σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 = π2py2 σ2pz2  Dimagnetic 
O_{2}  8 + 8 = 16  σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px1 = π*2py1  Paramagnetic 
B_{2}  5 + 5 = 10  σ1s2 σ*1s2 σ2s2 σ*2s2 π2px1 = π2py1  Paramagnetic 
NO  7 + 8 = 15  σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 = π2py2 π*2px1 = π*2py0  Paramagnetic 
Hence, the correct answer is option (1).
Question 73:
On treatment of 100 mL of 0.1 M solution of CoCl_{3}.6H_{2}O with excess AgNO_{3}; 1.2 × 10^{22} ions are precipitated. The complex is:
Option 1:  [Co(H_{2}O)_{3}Cl_{3}].3H_{2}O 
Option 2:  [Co(H_{2}O)_{6}]Cl_{3} 
Option 3:  [Co(H_{2}O)_{5}Cl]Cl_{2}.H_{2}O 
Option 4:  [Co(H_{2}O)_{4}Cl_{2}]Cl.2H_{2}O 
Solutions:
CoCl3.6H2O + AgNO3 → nAgCl↓1.2 × 10226.0 × 1023 = 2 × 102 mole = 0.020.01 × n = 0.02n = 2Therefore, the complex is [Co(H_{2}O)_{5}Cl]Cl_{2}.H_{2}O.
Hence, the correct answer is option (3).
Question 74:
pK_{a} of a weak acid (HA) and pK_{b} of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is :
Option 1:  6.9 
Option 2:  7.0 
Option 3:  1.0 
Option 4:  7.2 
Solutions:Given: pH of weak acid, HA = 3.2 pH of weak base, BOH = 3.4
To find: pH of salt AB = ?
pH of the solution of salt AB formed by weak acid HA and weak acid BOH is given as :
pH=12pKw – pKb + pKapH = 1214 – 3.4 + 3.2pH = 6.9Hence, the correct answer is option (1).
Question 75:
The increasing order of the reactivity of the following halides for the S_{N}1 reaction is :
Option 1:  (II) < (I) < (III) 
Option 2:  (I) < (III) < (II) 
Option 3:  (II) < (III) < (I) 
Option 4:  (III) < (II) < (I) 
Solutions:Order of reactivity for S_{N}1 mechanism is in accordance with order of stability of carbocation involved:
Hence, the correct answer is option (1).
Question 76:
Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is :
Option 1:  both form soluble bicarbonates 
Option 2:  both form nitrides 
Option 3:  nitrates of both Li and Mg yield NO_{2} and O_{2} on heating 
Option 4:  both form basic carbonates 
Solutions:Magnesium form basic carbonate [MgCO_{3}.Mg(OH)_{2}], while lithium form normal carbonate (Li_{2}CO_{3}). So, the given statement in option 4, both form basic carbonate, is incorrect.
Hence, the correct answer is option (4).
Question 77:
The correct sequence of reagents for the following conversion will be :
Option 1:  CH_{3}MgBr, H^{+}/CH_{3}OH, [Ag(NH_{3})_{2}]^{+}OH^{–} 
Option 2:  CH_{3}MgBr, [Ag(NH_{3})_{2}]^{+}OH^{–}, H^{+}/CH_{3}OH 
Option 3:  [Ag(NH_{3})_{2}]^{+}OH^{–}, CH_{3}MgBr, H^{+}/CH_{3}OH 
Option 4:  [Ag(NH_{3})_{2}]^{+}OH^{–}, H^{+}/CH_{3}OH, CH_{3}MgBr 
Solutions:
Hence, the correct answer is option (4).
Question 78:
The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are :
Option 1:  ClO2 and ClO3 
Option 2:  Cl– and ClO– 
Option 3:  Cl– and ClO2 
Option 4:  ClO– and ClO3 
Solutions:Complete chemical equation is:
2 NaOHCold and dilute + Cl2 → NaCl + NaOCl + H2OOr2 NaOH + Cl2 → Cl + ClO + Na+ + H2OHence, the correct answer is option (2).
Question 79:
Which of the following compounds will form significant amount of meta product during mononitration reaction ?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Among the given compounds, aniline will give significant amount of meta product during mononitration reaction. Because aniline in acidic medium converts into anilinium ion and it is meta directing, therefore, significant amount of meta product is obtained.
Hence, the correct answer is option (2).
Question 80:
3Methylpent2ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is:
Option 1:  Zero 
Option 2:  Two 
Option 3:  Four 
Option 4:  Six 
Solutions: Total number of stereo centres present in the product of 2bromo3methylpentane is two. Number of stereoisomers = 2^{n} = 4 [n = Number of chiral carbon atom]
Hence, the correct answer is option (3).
Question 81:
Two reactions R_{1} and R_{2} have identical preexponential factors. Activation energy of R_{1} exceeds that of R_{2} by 10kJ mol^{–1}. If k_{1} and k_{2} are rate constants for reactions R_{1} and R_{2} respectively at 300 K, then ln(k_{2}/k_{1}) is equal to : (R = 8.314 J mol^{–1} K^{–1})
Option 1:  12 
Option 2:  6 
Option 3:  4 
Option 4:  8 
Solutions:Given:
k1 = AeEa1/RTk2 = AeEa1 – 10/RTSince,lnk2k1 = 10RT = 108.314 × 103 × 300 =4.0093 ≈4Hence, the correct answer is option (3).
Question 82:
Which of the following molecules is least resonance stabilized?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Given molecules in options 1, 2 and 4 are aromatic and stablized by according to Huckel’s rule. Molecule in option 3 is nonaromatic, hence it is least resonance stablised.
Hence, the correct answer is option (3).
Question 83:
The group having isoelectronic species is:
Option 1:  O^{−}, F^{−}, Na, Mg^{+} 
Option 2:  O^{2−}, F^{−}, Na, Mg^{2+} 
Option 3:  O^{−}, F^{−}, Na^{+}, Mg^{2+} 
Option 4:  O^{2−}, F^{−}, Na^{+}, Mg^{2+} 
Solutions:Isoelectronic species are those which have same number of electrons. Among the given group of species O^{2}, F^{–}, Na^{+} and Mg^{2+} have same number of electrons (all contain 10 electrons).
Hence, the correct answer is option (4).
Question 84:
The radius of the second Bohr orbit for hydrogen atom is :
(Planck’s Const. h = 6.6262 × 10^{–34} Js; mass of electron = 9.1091 × 10^{–31} kg; charge of electron e = 1.60210 × 10^{–19} C; permittivity of vacuum ∈_{0} = 8.854185 × 10^{–12} kg^{–1}m^{–3}A^{2})
Option 1:  4.76 Å 
Option 2:  0.529 Å 
Option 3:  2.12 Å 
Option 4:  1.65 Å 
Solutions:The radius expression for the second Bohr orbit for hydrogen atom is
rn = 52.9 n2Z pm = 0.529 n2Z A∘.
rn = 0.529 221 A∘ = 2.116 A∘ ≈ 2.12 A∘Hence, the correct answer is option (3).
Question 85:
The major product obtained in the following reaction is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions: DIBALH reduce both ester and carboxylic acid into aldehyde at low temperature.
Hence, the correct answer is option (1).
Question 86:
Which of the following reactions is an example of a redox reaction ?
Option 1:  XeF_{2} + PF_{5} → [XeF]^{+}
PF6 
Option 2:  XeF_{6} + H_{2}O → XeOF_{4} + 2HF 
Option 3:  XeF_{6} + 2H_{2}O → XeO_{2}F_{2} + 4HF 
Option 4:  XeF_{4} + O_{2}F_{2} → XeF_{6} + O_{2} 
Solutions:
Xe+4F4 + O2+1F2 → Xe+6F6 + O02In the given reaction, oxidation of XeF_{4} and reduction of O_{2}F_{2} occurs, hence, this is a kind of redox reaction.
Hence, the correct answer is option (4).
Question 87:
A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be :
Option 1:  22a 
Option 2:  2a 
Option 3:  a2 
Option 4:  2a 
Solutions:In face centred cubic (fcc), atoms are in close contact along face diagonal of fcc unit cell. Hence,
4r = 2a2r = 2a2 = a2Hence, the correct answer is option (3).
Question 88:
Sodium salt of an organic acid ‘X’ produces effervescence with conc. H_{2}SO_{4}. ‘X’ reacts with the acidified aqueous CaCl_{2} solution to give a white precipitate which decolourises acidic solution of KMnO_{4}. ‘X’ is :
Option 1:  HCOONa 
Option 2:  CH_{3}COONa 
Option 3:  Na_{2}C_{2}O_{4} 
Option 4:  C_{6}H_{5}COONa 
Solutions:Chemical reactions:
Sodium salt of organic acid (Na2C2O4) produces effervescence with conc. H2SO4.Na2C2O4 + H2SO4 → Na2SO4 + H2O + CO + CO2Na2C2O4 reacts with acidified aqueous CaCl2 to give white precipitate CaC2O4.Na2C2O4 + CaCl2 → CaC2O4 ↓ + 2 NaClWhite precipitate CaC2O4 decolourises acidic solution of KMnO4.8 H2SO4 + 5 CaC2O4 + 2 KMnO4 → 5 CaSO4 + K2SO4 + 2 MnSO4 + 8 H2O + 10 CO2Hence, the correct answer is option (3).
Question 89:
A water sample has ppm level concentration of following anions
F^{–} = 10 ;
SO42= 100;
NO3=50The anion/anions that make/makes the water sample unsuitable for drinking is/are :
Option 1:  both
SO42and NO3 
Option 2:  only F^{–} 
Option 3:  only
SO42 
Option 4:  only
NO3 
Solutions:Internation standards for drinking water Fluoride ion (F) concentration upto 1 ppm Nitrate ion (NO_{3}^{–}) concentration upto 50 ppm Sulphate ion (SO_{4}^{2}) concentration upto 500 pmm From the given data, it can be conclude that only fluoride ion make the water sample unsuitable for drinking.
Hence, the correct answer is option (2).
Question 90:
Which of the following, upon treatment with tertBuONa followed by addition of bromine water, fails to decolourize the colour of bromine
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Hence, the correct answer is option (4).