## The side of a cube is measured by vernier calipers (10 divisions of the vernier scale coincide with 9 divisions of the main scale…

Q: The side of a cube is measured by vernier calipers (10 divisions of the vernier scale coincide with 9 divisions of the 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.

Sol. Least count of vernier calipers =(1 division of main scale )/(Number of division in vernier scale)

= 1/10 = 0.1 mm

The side of cube = 10 mm + 1 × 0.1 mm = 1.01 cm

Now,density = Mass/Volume

$\large = \frac{2.736}{(1.01)^3 cm^3}$

= 2.66 g cm-3

## n Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm…

Q: In Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. the length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of young’s modulus of the material of the wire from these data.

Sol. Maximum percentage error in Y is given by

$\large Y = \frac{W}{\frac{\pi D^2}{4}} \times \frac{L}{x}$

$\large \frac{\Delta Y}{Y} = 2 \frac{\Delta D}{D} + \frac{\Delta L}{L} + \frac{\Delta x}{x}$

$\large \frac{\Delta Y}{Y} = 2 \frac{0.001}{0.05} + \frac{0.1}{110} + \frac{0.001}{0.125}$

= 0.0489

## 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…

Q: 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 1 mm and 47th circular division coincides with the main scale. Find the curved surface area of the wire in cm2 to appropriate significant figures. (Use π = 22/7)

Sol. Least Count = (1 mm)/100 = 0.01 mm

Diameter= MSR + CSR(LC)

=1 mm + 47(0.01)mm =1.47 mm

Surface area = π Dl = 22/7 × 1.47 × 56 mm2

=2.58724 cm2 = 26 cm2

## If c is the velocity of light, h is Planck’s constant and G is Gravitational constant are taken as fundamental quantities…

Q: If c is the velocity of light, h is Planck’s constant and G is Gravitational constant are taken as fundamental quantities. Them the dimensional formula of mass is

Sol. c = [LT-1 ] …(i)

h = [ML2 T-1 ] …(ii)

G = [M-1 L3 T-2 ] …(iii)

Solving (ii) and (iii)

h/G = [(ML2 T-1)/(M-1 L3 T-2 )]

=[M2 L-1 T1 ]

Substituting (i) in above

h/G = M2/c

⇒ [M] = [h1/2 G-1/2 c1/2 ]

## Derive an expression for the time period of a simple pendulum of mass (m), length (l) at a place where acceleration due to gravity is (g)

Q: Derive an expression for the time period of a simple pendulum of mass (m), length (l) at a place where acceleration due to gravity is (g).

Sol. Let the time period of a simple pendulum depend upon the mass of bob m. length of pendulum l, and acceleration due to gravity g, then

t ∝ ma lb gc

⇒ t = k ma lb gc

M0 L0 T1 = Ma Lb [LT-2]c

⇒ M0 L0 T1 = Ma Lb+c T-2c

Comparing the powers of M, L, and T on both sides, we get a = 0, b =1/2 and c = -1/2. Putting these values,

⇒ a = 0,b = 1/2 and c = -1/2.Putting these values,

$\large T = k\sqrt{\frac{l}{g}}$

which is the required relation.