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11th Standard English Medium Physics Subject Work, Energy and Power Book Back 1 Mark Questions with Solution Part - I updated Book back Questions

Published on: 13/05/2022

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Std: 11th Standard - Physics

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Questions:

Multiple Choice Question

5 x 1 = 5

Two equal masses m₁ and m₂ moving along the same straight line with velocities +3 m/s and -5 m/s respectively collide elastically. Their velocities after the collision will be respectively _______________.

(a)

- 4 m/s and +4 m/s

(b)

+4 m/s for both

(c)

- 3 m/s and +5 m/s

(d)

- 5 m/s and + 3 m/s

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of

(a)

\(\frac{2}{3}\)k

(b)

\(\frac{3}{2}\)k

(c)

(d)

A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). If U (0) = 0, the graph of U(x) versus x will be (where U, is the potential , energy function)

(a)

(b)

(c)

(d)

If the potential energy of the particle is \(\alpha -\frac { \beta }{ 2 } { x }^{ 2 }\), then force experienced by the particle is

(a)

F = \(\frac { \beta }{ 2 } { x }^{ 2 }\)

(b)

F = βx

(c)

F = -βx

(d)

F = -\(\frac { \beta }{ 2 } { x }^{ 2 }\)

The work done by the conservative force for a closed path is

(a)

always negative

(b)

zero

(c)

always positive

(d)

not defined

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Answers Key:

Multiple Choice Question

5 x 1 = 5

(d)
- 5 m/s and + 3 m/s

For any spring kl = constant
Length of the longer piece
\(=\frac{2 l}{3} \)
\(\therefore k^{1} \times \frac{2 l}{3} =k l \)
\(\therefore k^{1}=\frac{k l \times 3}{2 l}=\frac{3}{2} k \)
\(\therefore k^{1}=\frac{3}{2} k \)

For a conservative force
\(F=-\frac{d v}{d t} \)
\(\int_{0}^{u(x)} d v=-\int_{0}^{x} F d x=-\int_{0}^{x} k x d x \)
\(\text { As } v(0)=0\)
\(U(x)=-\frac{k x^{2}}{2}\)
Thus, the graph of U(x) versus (x) will be a parabola, symmetric about U - ax is bying below x - ax is with its vertex at the origin. Hence the correct answer is C

\(\text {Potential energy } P . E=\alpha-\frac{\beta}{2} x^{2}\)
P.E = Work = Fx
\(P=\alpha-\frac{\beta}{2} x^{2}\)
\(\text {Force }=\frac{d p}{d x}=\frac{d}{d x}\left(\alpha-\frac{\beta}{2} x^{2}\right)\)
\(=0-\frac{\beta}{2} \times 2 x =-\beta x \)