Practice free Mechanical Properties of Solids (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Calculate the work done, if wire is loaded by 'M g' weight and the increase in length is 'l' ?
$ work done = { 1 \over 2 } Fl = { Mgl\ove r2 } $
Two wires of same diameter of the same material having the length l and 2 l . If the force F is applied on each, what will be the ratio of the work done in the two wires ?
$ w = { 1 \over 2} { (stress)^2 \over Y } \times volume $ $ As F, Aand Y are same - W \alpha volume (area is same)$ $ w \alpha l $ (v =A1) $ { w_1 \over w_2 } = { l_1 \over l_2 } = { l_1 \over 2 l } = {1 \over 2} $
A 5 meter long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 meter above the four. The wire was elongated by 1 mm. What is the stored in the wire due to stretching ?
$ { w = {1 \over 2 } \times F \times l $
If the force constant of a wire is k. What is the work done in increasing the length of the wire by l ?
$ K = {F \over l } and w = { 1 \over 2 } F l = {1 \over 2 } kl \times l = {1 \over 2 } kl^2 $
Wire A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each is stretched by the same tension, what is the ratio of energy stored in A to that in B ?
$ U = { 1 \over 2} Fl = { F^2 l \over 2 AY } ; U \alpha {L \over r^2 } (F and Y are constant ) $
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower and. The weight stretches the wire by 1 mm. Then what is the elastic energy stored in the wire ?
$ U = {1 \over 2 } Fl $
A brass rod of cross sectional area 1 cm2 and length 0.2 mis compressed length wise by a weight of 5 kg. If young's modulus of elasticity of brass is $ 1 \times 10 ^ {11} N /m^2 $ and $ g = 10 m/s^2 $ . Then what will be increase in the energy of rod ?
$ U = {1 \over 2} \times {(stress )^2 \over Y } \times volume $
The work per unit volume to stretch the length by 1% of a wirewith cross - sectional area $1 mm^2$ will be ………..$( Y = 9 \times 10 ^{11} N/m^2 ) $
$ U = {1 \over 2 } \times YX (strain )^2 $
A wire of length 50 cm and cross - sectional area of 1 mm2 is extended by $1mm^2 $ what will be the required work ? $ ( Y = 2 \times 10 ^{11} Nm^{-2} $
$ w = { YAl^2 \over 2 L } $
If a spring extends by x cm loading then what is the energy stored by the spring ? (If T is tensionin the spring & K is spring constant)
$ U = {F^2 \over 2K } = { T^2 \over 2K } $
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