The sketch between cause (magnitude of acceleration) and effect (magnitude of displacement) is a straight line.

As it moves with respect to x axis
\(\Delta p_{x}=-m u, \Delta p_{y}=0\)

\(\text { Stress, } \ \frac{F}{A}=Y \frac{\Delta L}{L}\)
\(\text { Stress } \mathrm{S}=\frac{\text { Force }}{\pi r^{2}}\)
\(S_{1} \propto \frac{1}{r_{1}^{2}} \ S_{2} \propto \frac{1}{r_{2}^{2}} \ F \text { is same }\)
\(\frac{S_{1}}{S_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}} \quad r_{1}=3 r_{2}\)
\(\text { Let } S_{1}=S_{x} \text { and } S_{2}=S_{y}\)
\( \frac{S_{x}}{S_{y}}=\frac{r_{2}^{2}}{\left(3 r_{2}\right)^{2}}=\frac{r_{2}^{2}}{9 r_{2}^{2}}=\frac{1}{9} \)
\( \frac{S_{x}}{S_{y}}=\frac{1}{9} \)
\(\therefore S_{y}=9 S_{x} \)

\(\text { Force } \overrightarrow{\mathbf{F}}=(2 i+\vec{j}) N\)
\(\text { Displacement } d=(5 \vec{i}+3 \vec{j})-(3 \vec{j}+\vec{k})\)
\(=(5 i-k) m\)
\(\text { Work done } W=F . d\)
\(=(2 \vec{i}+\vec{j})(5 i-k)\)
\(=10-0-0=10 J \)

Answers (a), (b) and (c) are all anticlockwiseand answer (d) alone is in the clockwise direction

Unit of a (Planck's constant) - Js
Unit of c (Velocity of light) - ms^-1
Unit of G (Gravitational Constant) - \(\frac{\mathrm{Nm}^{2}}{\mathrm{Kg}^{2}}\)
\(\therefore \text { Unit of } \frac{h c}{G} \text { is }=\frac{J s \times m s^{-1}}{N m^{2} / k g^{2}} \)
\(=\frac{N m s \times m s^{-1} \times k g^{2}}{N m^{2}}[J=N m] =\mathrm{kg}^{2}\)