1

IIT-JEE 2005

Subjective
The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures.

2.66 gm/cm3
2

IIT-JEE 2004

Subjective
A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the diameter of a wire of length 5.6 cm. The main scale reading is 1mm and 47th circular division coincides with the main scale. Find the curved surface area of wire in cm2 to appropriate significant figures. ( use $\pi = {{22} \over 7}$ )

2.6 cm2

Explanation

The distance moved on the linear scale when circular scale makes one complete rotation is p = 1 mm (pitch). The number of divisions on the circular scale is N = 100. Thus, one division on the circular scale is LC = ${p \over N} = {1 \over {100}}$ = 0.01 mm. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR) is 47. Thus, the diameter of the wire is

d = LSR + CSR $\times$ LC

= 1 + 47 $\times$ 0.01 = 1.47 mm = 0.147 cm.

The curved surface area is S = 2$\pi$rL. That is,

$S = 2\pi \left( {{d \over 2}} \right)L$

$S = \pi dL = \pi \left( {{{1.47} \over {10}}} \right)5.6 =$ 2.5848 cm2 = 2.6 cm2

which is corrected to two significant digits.
3

IIT-JEE 2004

Subjective
In Searle's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm ( measured by a scale of least count 0.001 cm ) and length is L = 110 cm ( measured by a scale of least count 0.1 cm ). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm ). Find maximum possible error in the value of Young's modulus. Screw gauge and and meter scale are free from error.

1.09 $\times$ 1010 N/m2
4

IIT-JEE 2003

Subjective
If nth divisions of the main scale coincide with (n+1)th divisions of vernier scale. Given one main scale division is equal to 'a' units. Find the least count of the vernier.

${a \over {n + 1}}units$

Explanation

We know that (n + 1) divisions on the vernier scale = n division on the main scale

Therefore,

1 division on the vernier scale = $\left( {{n \over {n + 1}}} \right)$ divisions on the main scale = $\left( {{n \over {n + 1}}} \right)a$ units

Therefore, the least count (LC) of the vernier caliper is

$1(MSD) - 1(VSD) = a - \left( {{n \over {n + 1}}} \right)a = {a \over {n + 1}}$