JEE Physics Motion In A Straight Line MCQs Set 27

Answer: (b) 2.0 m/s

Explanation:
1. Displacement AB is the diameter \( =2R=2 \) m.
2. Average velocity \( =\) displacement/time \( =2/1=2 \) m/s (the path length \( \pi R \) is not used).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (c) \( \tfrac1{\sqrt2} \) m/s² towards North-West

Explanation:
1. \( \Delta\vec v=\vec v_f-\vec v_i=5\hat j-5\hat i \), of magnitude \( 5\sqrt2 \) m/s towards North-West.
2. \( a_{av}=5\sqrt2/10=1/\sqrt2 \) m/s² towards North-West.

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (a) \( 2R,\ \pi R \)

Explanation:
1. In half a revolution the particle goes from one end of a diameter to the other: displacement \( =2R \).
2. Distance \( =\) half the circumference \( =\pi R \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (b) \( \dfrac{b^2\tau}{2} \)

Explanation:
1. \( a=v\,dv/dx=b\sqrt x\cdot\dfrac{b}{2\sqrt x}=\dfrac{b^2}{2} \), a constant.
2. Starting from rest at the origin, \( v=at=\dfrac{b^2\tau}{2} \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (b) \( \dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}\,v_0\right) \)

Explanation:
1. Going up, \( \dfrac{dv}{dt}=-(g+\gamma v^2) \), so \( t=\int_0^{v_0}\dfrac{dv}{g+\gamma v^2} \).
2. \( =\dfrac{1}{\sqrt{\gamma g}}\tan^{-1}\left(\sqrt{\dfrac{\gamma}{g}}\,v_0\right) \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (d) \( \tan^{-1}(0.5) \)

Explanation:
1. \( T\cos\theta=mg \) and \( T\sin\theta=qE \), so \( \tan\theta=qE/mg \).
2. \( qE=5\times10^{-6}\times2000=10^{-2} \) N and \( mg=2\times10^{-3}\times10=2\times10^{-2} \) N, so \( \tan\theta=0.5 \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (b) \( 20\sqrt2 \) m

Explanation:
1. x: \( 2+5(2)+\tfrac12(4)(2)^2=20 \) m.
2. y: \( 4+4(2)+\tfrac12(4)(2)^2=20 \) m.
3. Distance \( =\sqrt{20^2+20^2}=20\sqrt2 \) m.

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (c) \( \sqrt{a_1a_2}\,t \)

Explanation:
1. Let A take \( t_0 \) and B take \( t_0+t \). Equal distances: \( \tfrac12a_1t_0^2=\tfrac12a_2(t_0+t)^2 \), so \( \sqrt{a_1}\,t_0=\sqrt{a_2}(t_0+t) \).
2. Hence \( t_0=\dfrac{\sqrt{a_2}\,t}{\sqrt{a_1}-\sqrt{a_2}} \).
3. \( v'=a_1t_0-a_2(t_0+t)=\sqrt{a_1}\,s-\sqrt{a_2}\,s \) with \( s=\sqrt{a_1}\,t_0 \), giving \( v'=\sqrt{a_1a_2}\,t \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (c) \( 2gH=nu^2(n-2) \)

Explanation:
1. Time to the top \( =u/g \), so total time \( T=nu/g \).
2. Taking upward as positive, \( -H=uT-\tfrac12gT^2=\dfrac{nu^2}{g}-\dfrac{n^2u^2}{2g} \).
3. \( 2gH=u^2n(n-2) \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.

Answer: (c) \( \dfrac{2n-1}{2n+1} \)

Explanation:
1. Distance in the nth second \( s_n=\tfrac12a(2n-1) \).
2. \( \dfrac{s_n}{s_{n+1}}=\dfrac{2n-1}{2n+1} \).

Teacher's Note:
1. Draw the situation and write the given quantities with signs before substituting.
2. Check the final answer for units and a sensible magnitude.