NEETprep Bank NEET Physics Mechanical Properties of Solids 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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Concept Questions :-

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The Young's modulus of a wire of length L and radius r is Y N/m2. If the length and radius are reduced to L/2 and r/2, then its Young's modulus will be -

(a) 1:1                           (b) 1:4

(c) 1:8                             (d) 8:1

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

(b) Young's modulus of wire does not varies with dimension of wire. It is the property of given material.

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When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

(a) 0.5 cm                              (b) 2 cm

(c) 4 cm                                 (d) 8 cm

Concept Videos :-

Concept Questions :-

(c)

      l=FLAY11r2(F,L and Y are constant)l2l1=r1r22=22=4l2=4l1=4cm

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 The area of cross-section of a wire of length 1.1 metre is 1mm2.It is loaded with 1 kg. If Young's modulus of copper is 1.1×1011N/m2, then the increase in length will be (If g=10m/s2)

(a) 0.01 mm                (b) 0.075 mm

(c) 0.1 mm                  (d) 0.15 mm

Concept Videos :-

Concept Questions :-

(c) l=mgLAY=1×10×1.11.1×1011×10-6m=0.1mm

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