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23P1Q10

10        A

Change in momentum per collision, \displaystyle \Delta p=2mv

Time interval between collisions for each molecule, \displaystyle \Delta t=\frac{2p}{v}

Force produced by each molecule, \displaystyle f=\frac{\Delta p}{\Delta t}=2mv\div \frac{2p}{v}=\frac{m{{v}^{2}}}{p}

Total average force, \displaystyle F=N\times f=N\times \frac{m{{v}^{2}}}{p}

COMMENT: It is unfortunate that p, which usual denotes momentum, is also the box’s length in this question. But you should be smart enough to not be confused by the setter.

23P1Q09

9          C

Options A, B and D are possible paths even if the gravitational force (towards the centre of Earth) is the only force acting on the spacecraft.

Option C is not possible. With the Earth continuously pulling the spacecraft towards the centre of the Earth, it is not possible for the spacecraft to travel along a straight path. The spacecraft must fire its rockets to at least counter the “leftward” component of the gravitational pull.

23P1Q05

5          B

Method 1

The overall C.G. lies somewhere between the C.G. of the top half and the bottom half.

Since the top half has twice the mass as the bottom half, the resultant C.G. should lie twice as near C.G. of the top half as the bottom. Meaning the C.G. is \displaystyle \frac{1}{3}\times 4\text{ cm}and \displaystyle \frac{2}{3}\times 4\text{ cm}from the C.G. of the top and bottom halves respectively.

From P, the distance would be \displaystyle 2+\frac{2}{3}(4)=4.7\text{ cm}

Method 2

Let the mass of the top half be 2m so the bottom half would be m.

Let the distance between the C.G. and P be x.

\displaystyle \begin{aligned}  & 2m\times 6.0+m\times 2.0=3m\times x \\ & x=4.67\text{ cm} \end{aligned}