# JEE Advanced 2016 Paper 2 (Code-6)

### Test Name: JEE Advanced 2016 Paper 2 (Code-6)

#### Question 1:

There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2, respectively, are

 Option 1: 2.87 and 2.86 Option 2: 2.85 and 2.82 Option 3: 2.87 and 2.87 Option 4: 2.87 and 2.83

Solutions:Least count of C1 =

1 MSD -1VSD=1-910MSD=110×1 mm=0.01 cmReading oon C1 = MSD + (VSD

×LC) = 2.8 + 7

×0.01 = 2.87

Since the length of VSD is greater that MSD, we have to find the reading from the basics. For C2, 1 VSD =

1110 mm=0.11 cm Distance measured from main scale = Distance measured from vernier scale

⇒2.8 cm+(8×0.1 cm)=(2.8 cm+x)+(7×0.11 cm)⇒x=0.03 cmTherefore, reading on C2 = 2.8 + 0.03 = 2.83 cm

Hence, the correct answer is option D.