
- A$\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} / \mathrm{s}^2$
- B$\frac{-\pi^2}{32} \mathrm{~cm} / \mathrm{s}^2$
- C$\frac{\pi^2}{32} \mathrm{~cm} / \mathrm{s}^2$
- ✓$-\frac{\sqrt{3}}{32} \pi^2 \mathrm{~cm} / \mathrm{s}^2$

At, $t =0$, the particle is at $x =0$.
Hence, the particle is starting $SHM$ from its mean position.
The equation of $SHM$ of particle will be, $x=A \sin \omega t$ $\qquad$
From the graph, one oscillation has been completed in time, $T =8 s$ and maximum displacement, $A=1 cm$
So, angular frequency, $\omega=\frac{2 \pi}{T}=\frac{2 \pi}{8}=\frac{\pi}{4} rad / s$
On putting above data in $(1)$
$x=1 \sin \left(\frac{\pi t}{4}\right)$
At, $t=\frac{4}{3} s$
$x=1 \sin \left(\frac{\pi}{4} \times \frac{4}{3}\right)=\frac{\sqrt{3}}{2} cm$
Now,
Acceleration of the particle at $t=\frac{4}{3} s$,
$a=-\omega^2 x=-\left(\frac{\pi}{4}\right)^2 \times \frac{\sqrt{3}}{2}$
$\Rightarrow a=-\frac{\sqrt{3} \pi^2}{32} cm / s ^2$
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$(a)$ The distance between the objective and eye piece is $20.02\; m$
$(b)$ The magnification of the telescope is $1000$
$(c)$ The image of the planet is erect and diminished
$(d)$ The aperture of eye piece is smaller than that of objective
The correct statements are :

| List-$I$ | List-$II$ |
| $P.$ $\quad Q _1, Q _2, Q _3, Q _4$, all positive | $1.\quad$ $+ x$ |
| $Q.$ $\quad Q_1, Q_2$ positive $Q_3, Q_4$ negative | $2.\quad$ $-x$ |
| $R.$ $\quad Q_1, Q_4$ positive $Q_2, Q_3$ negative | $3.\quad$ $+ y$ |
| $S.$ $\quad Q_1, Q_3$ positive $Q_2, Q_4$ negative | $4.\quad$ $-y$ |

| List-$I$ | List-$II$ |
| $(A)$ A force thatrestores anelastic body of unit area to its original state | $(I)$ Bulkmodulus |
| $(B)$ Two equal andopposite forcesparallel toopposite faces | $(II)$Young'smodulus |
| $(C)$Forcesperpendiculareverywhere tothe surface perunit areasameeverywhere | $(III)$ Stress |
| $(D)$Two equal andopposite forceperpendicular toopposite faces | $(IV)$ Shearmodulus |
Choose the correct answer from the options given below: