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1

### JEE Main 2021 (Online) 31st August Evening Shift

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $${1 \over 4}$$ times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-
A
T
B
$${3 \over 2}$$T
C
$${3 \over 4}$$T
D
$${4 \over 3}$$T

## Explanation

$$T = 2\pi \sqrt {l/g}$$

When bob is immersed in liquid

mgeff = mg $$-$$ Buoyant force

mgeff = mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)

$$= mg - v{\rho \over 4}g$$

$$= mg - {{mg} \over 4} = {{3mg} \over 4}$$

$$\therefore$$ $${g_{eff}} = {{3g} \over 4}$$

$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}}$$

$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$

By solving

$${T_1} = {4 \over 3}2\pi \sqrt {l/g}$$

$${T_1} = {{4T} \over 3}$$
2

### JEE Main 2021 (Online) 31st August Evening Shift

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 $$\times$$ 103 kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 $$\times$$ 1011 Pa, g = 9.8 m/s2]
A
3.60 $$\times$$ 10$$-$$8
B
2.60 $$\times$$ 10$$-$$7
C
1.87 $$\times$$ 10$$-$$3
D
7.07 $$\times$$ 10$$-$$4

## Explanation

Force on each column = $${{mg} \over 4}$$

Strain = $${{mg} \over {4AY}}$$

$$= {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$

= 2.6 $$\times$$ 10$$-$$7
3

### JEE Main 2021 (Online) 31st August Morning Shift

A uniform heavy rod of weight 10 kg ms$$-$$2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 $$\times$$ 1011 Nm$$-$$2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.
A
2 $$\times$$ 10$$-$$9 m
B
5 $$\times$$ 10$$-$$8 m
C
4 $$\times$$ 10$$-$$8 m
D
5 $$\times$$ 10$$-$$10 m

## Explanation

We know,

$$\Delta l = {{WL} \over {2AY}}$$

$$\Delta l = {{10 \times 1} \over {2 \times 5}} \times 100 \times {10^{ - 4}} \times 2 \times {10^{11}}$$

$$\Delta l = {1 \over 2} \times {10^{ - 9}} = 5 \times {10^{ - 10}}$$ m

Option (d)
4

### JEE Main 2021 (Online) 27th August Morning Shift

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 $$\times$$ 10$$-$$5 m and density 1.2 $$\times$$ 103 kgm$$-$$3 ? Take viscosity of liquid = 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2. (Neglect buoyancy due to air).
A
3.8 $$\times$$ 10$$-$$11 N
B
3.9 $$\times$$ 10$$-$$10 N
C
1.8 $$\times$$ 10$$-$$10 N
D
5.8 $$\times$$ 10$$-$$10 N

## Explanation

Viscous force = Weight

$$= \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g$$

= 3.9 $$\times$$ 10$$-$$10

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