NEET Practice Questions (MCQs) with Answers & Solutions

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A steel wire of 1 m long and  cross section area 1 mm2 is hang from rigid end. When mass of 1kg is hung from it then change in length will be (given Y=2×1011N/m2)

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Explanation

(c) l=MgLYA=1×10×12×1011×10-6=0.05mm

A force F is applied on the wire of radius r and length L and change in the length of wire is l.  If the same force F is applied on the wire of the same material and radius 2r and length 2L, Then the change in length of the other wire is

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Explanation

(c)

 l=FLAYlLr2       F and Y are constantl2l1=L2L1×r1r22=2×122=12l2=l12

i.e., the change in the length of other wire is l2

An iron rod of length 2m and cross section area of 50 X 10-6 m2 , is stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of the iron rod is-

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Explanation

(a) Y=MgLAl=250×9.8×250×10-6×0.5×10-3                 =19.6×1010N/m2

In which case there is maximum extension in the wire, if same force is applied on each wire

 L = 400 cm, d = 0.01 mm

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Explanation

(d) lLr2                                 Y and F are constant

Maximum extension takes place in that wire for which the ratio of Lr2 will be maximum.

The extension of a wire by the application of load is 3 mm. The extension in a wire of the same material and length but half the radius by the same load is -

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Explanation

(a) l=FLAYl1r2    F,L and Y are constantl2l1=r1r22=22l2=4l1=4×3=12mm

The isothermal elasticity of a gas is equal to

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Explanation

(c) Isothermal elasticity Ki=P

The adiabatic elasticity of a gas is equal to

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Explanation

(c) Adiabatic elasticity Kα=γP

The specific heat at constant pressure and at constant volume for an ideal gas are Cp and Cv and its adiabatic and isothermal elasticities are E andEθ  respectively. The  ratio of E to Eθ is

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Explanation

(b) Ratio of adiabatic and isothermal elasticities

EEθ=γPP=γ=CpC

The compressibility of water is 4×10-5 per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be -

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Explanation

(a) C=1K=V/VPV=C×P×V=4×10-5×100×100=0.4cc

If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1%. If the density of the water is 1×103kg/m3 and g=10m/s2, then the volume elasticity in  will be

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Explanation

(d) K=PV/V=hpgV/V=200×103×100.1/100=2×109 

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