- A$0.4\ m$ from mass of $ 0.3\ kg$
- ✓$0.98\ m$ from mass of $0.3\ kg$
- C$0.70\ m$ from mass of $0.7\ kg$
- D$0.98\ m$ from mass of $0.7\ kg$
Given$:$ $w$ is constant
So required work done to be minimum implies that $I$ must be minimum.
Let the rotational axis passes through O. $I=0.3\left(x^{2}\right)+0.7(1.4-x)^{2}$
For $I$ to be minimum, $\quad \frac{d I}{d x}=0$
$\Longrightarrow 0.3 \times 2(x)-0.7 \times 2(1.4-x)=0$
$\Longrightarrow x=0.98 m$
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Statement $I$ : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_1\right)$
Statement $II$ : In ventury tube shown $2 \mathrm{gh}=v_1^2-v_2^2$
In the light of the above statements, choose the most appropriate answer from the options given below.

| column $I$ | column $II$ |
| $(A)$ $U _1( x )=\frac{ U _0}{2}\left[1-\left(\frac{ x }{ a }\right)^2\right]^2$ | $(P)$ The force acting on the particle is zero at $x = a$. |
| $(B)$ $U _2( x )=\frac{ U _0}{2}\left(\frac{ x }{ a }\right)^2$ | $(Q)$ The force acting on the particle is zero at $x=0$. |
| $(C)$ $U _3( x )=\frac{ U _0}{2}\left(\frac{ x }{ a }\right)^2 \exp \left[-\left(\frac{ x }{ a }\right)^2\right]$ | $(R)$ The force acting on the particle is zero at $x =- a$. |
| $(D)$ $U _4( x )=\frac{ U _0}{2}\left[\frac{ x }{ a }-\frac{1}{3}\left(\frac{ x }{ a }\right)^3\right]$ | $(S)$ The particle experiences an attractive force towards $x =0$ in the region $| x |< a$. |
| $(T)$ The particle with total energy $\frac{ U _0}{4}$ can oscillate about the point $x=-a$. |