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


[Only for Dropper and XII batch]

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(3)Differentiate VPn=C w.r.to V using multiplication rule:VddVPn+PnddVV=0VnPn-1dPdV+Pn.1=0VnPn-1dP+Pn.dV=0nVPn-1dP=-Pn.dVdPdV=-PnnVPn-1=-PnVBulk Modulus=-dPdVV=--PnVV=Pn

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Concept Videos :-

Concept Questions :-

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Concept Videos :-

#10 | Stress, Strain Curve
#11 | Application of Elastic Behaviour of Material

Concept Questions :-

Stress-strain curve

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Concept Videos :-

#1 | Potential Energy & Elastic Behaviour

Concept Questions :-

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Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

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Concept Videos :-

#9 | Poisson Ration

Concept Questions :-

Poisson ratio

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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)PB

(b) B3p

(c) 3pB

(d)p3B

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

(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=F/AV/V

      VV=PBVV=PB

Hence p is applied pressure on the object and VV is

volume strain 

Fractional decreasesin volume 

       VV=3RR               So, V=43πR3

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

Hence VV=3RR=PBRB=P3B

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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

Concept Videos :-

#2 | Stress
#3 | Strain
#4 | Hooks's Law & Young's Modules
#5 | Bulk Modules & Shear Modules
#6 | Solved Problems: Set 1
#7 | Extension in a Stretched Wire
#8 | Derivation for Energy Stored in a Wire
#12 | Solved Problems: Set 2
#13 | Solved Problems: Set 3

Concept Questions :-

Stress-strain

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

Concept Videos :-

Concept Questions :-

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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