NEET UG PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Easy

1 of 1

Application +4 | -1

Two forces

F_1

and

F_2

are applied on two rods

P

and

Q

of same materials such that elongation in rods are same. If ratio of their radii is

x

y

and ratio of length is

m: n

, then ratio of

F_1: F_2

View answer and explanation

Correct option: B

\left(\frac{x}{y}\right)^2 \cdot \frac{n}{m}

Option B is correct. The ratio of forces is derived from the Young's modulus formula, considering the same elongation and different radii and lengths of the rods.

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A 20 kg load is suspended by a wire of area of cross section is

1\,m{m^2}

. The stress produced in

N/{m^2}

\left[ {g = 10\,m{s^{ - 2}}\,} \right]\,

In magnitude hydraulic stress is equal to: A wire extends by

\mathrm{1 mm}

when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be Stress to strain ratio is equivalent to A

2\,\,m

long rod of radius

1\,\,cm

which is fixed from one end is given a twist of

0.8

radians. The shear strain developed will be The relationship between Young’s modulus

Y

, Bulk modulus

\mathrm{K}

and modulus of rigidity

\eta