# NEET and AIPMT NEET Physics Mechanical Properties of Solids MCQ Questions Solved

[Only for Dropper and XII batch]

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NEET - 2017

The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in radius is

(a)$\frac{P}{B}$

(b) $\frac{B}{3p}$

(c) $\frac{3p}{B}$

(d)$\frac{p}{3B}$

(d)The object is spherical and the bulk modulus is represented by B. It is the ratio of normal stress to the volumetric strain.

Hence             $B=\frac{F/A}{∆V/V}$

Hence p is applied pressure on the object and $\frac{∆v}{v}$ is

volume strain

Fractional decreasesin volume

Volume of the sphere decreases due to the decrease in its radius.

Hence $\frac{∆v}{v}=\frac{3∆R}{R}=\frac{P}{B}⇒\frac{∆R}{B}=\frac{P}{3B}$

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NEET - 2015

The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

(a)1:2

(b)2:1

(c)4:1

(d)1:1

Given, Ysteel=2Ybrass and Ls=Lb and As=Ab

such that ΔLs=ΔLb

As we know,Young's modulus

Y=stress/strain=W/A/ΔL/L

i.e. Ws/Wb=Ys/Yb=2Yb/Yb=2/1

=> 2:1

Thus, weight added to the steel and brass wires must be in the ratio of 2:1

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NEET - 2014

Copper of fixed volume v is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is  Δl. Which of the following graphs is a straight line?

(a)Δl versues 1/l

(b)Δl versus l2

(c)Δl versus 1/l2

(d) Δl versus l

For Hooke's law, Young Modulus

y=F/A/Δl/l

=F x l/Δl x A ...(i)

V = A x l=constant ...(ii)

From Eqs. (i) and (ii),

y=F x lxl / Δl x V

Δl=F/V x y x l2

=>Δl∝l2

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