Correct Answer is Option (D)

The relationship between the length of a pendulum and its period is given by the equation:

$T=2\pi \sqrt{\frac{L}{g}}$

$\Rightarrow$ ${\mathrm{T}}^2=\frac{4{\pi }^2}{g}\cdot \mathrm{L}$

where T is the period of the pendulum, g is the acceleration due to gravity, and L is the length of the pendulum. This equation shows that the square of the period is directly proportional to the length of the pendulum.

Therefore, a graph of T² vs. L will be a straight line with a positive slope, passing through the origin. The slope of the line will be equal to $\frac{4{\pi }^2}{g}$, and its intercept will be zero.

Note : Graph of T vs. L will be a parabola not T² vs. L as in equation of parabola y² = 4ax, graph of y vs. x is parabola not y² vs. x. As y² vs. x is a straight line.

Correct Answer is Option (C)

At surface of earth time period

$\mathrm{T}=2\pi \sqrt{\frac{\ell }{\mathrm{g}}}$

At height $\mathrm{h}=\mathrm{R}$

$\begin{array}{rl}& {\mathrm{g}}^{\prime }=\frac{\mathrm{g}}{{(1+\frac{\mathrm{h}}{\mathrm{R}})}^2}=\frac{\mathrm{g}}{4}\\ \\ & \therefore \mathrm{xT}=2\pi \sqrt{\frac{\ell }{(\mathrm{g}/4)}}\\ \\ & \Rightarrow \mathrm{xT}=2\times 2\pi \sqrt{\frac{\ell }{\mathrm{g}}}\\ \\ & \Rightarrow \mathrm{xT}=2\mathrm{T}\Rightarrow \mathrm{x}=2\end{array}$

Correct Answer is (99)

Given the amplitude $A$ and the length $l$ of the pendulum, we can find the maximum angular displacement ${\theta }_0$:

$\sin {\theta }_0=\frac{A}{l}=\frac{10}{100}=\frac{1}{10}$

By conservation of energy, the following equation holds:

$\frac{1}{2}m{v}^2=mgl(1-\cos \theta )$

The maximum tension occurs at the mean position (i.e., when the pendulum is vertical). At this point, we have:

$\begin{array}{rlr}& T-mg=\frac{m{v}^2}{l}& \Rightarrow T=mg+\frac{m{v}^2}{l}\end{array}$

Substituting the conservation of energy equation, we get:

$\begin{array}{rl}& T=mg+2mg(1-\cos \theta )\\ \\ & =mg[1+2(1-\sqrt{1-{\sin }^2\theta })]\\ \\ & =mg[3-2\sqrt{1-\frac{1}{100}}]\\ \\ & =\frac{250}{1000}\times 10[3-2(1-\frac{1}{200})]=\frac{99}{40}\\ \\ & \therefore x=99\end{array}$

So, the value of $x$ is 99.

Correct Answer is (5)

${\mathrm{mg}}^{\prime }=\mathrm{mg}-{\mathrm{F}}_{\mathrm{B}}$

${\mathrm{g}}^{\prime }=\frac{\mathrm{mg}-{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{F}}_{\mathrm{B}}}$

$=\frac{{\rho }_{\mathrm{B}}\mathrm{Vg}-{\rho }_{\mathrm{w}}\mathrm{Vg}}{{\rho }_{\mathrm{B}}\mathrm{V}}$

$=\left(\frac{{\rho }_{\mathrm{B}}-{\rho }_{\mathrm{w}}}{{\rho }_{\mathrm{B}}}\right)\mathrm{g}$

$=\frac{5-1}{5}\times \mathrm{g}$

$=\frac{4}{5}\mathrm{g}$

We know, $T=2\pi \sqrt{\frac{\ell }{\mathrm{g}}}$

$\frac{{\mathrm{T}}^{\prime }}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{{\mathrm{g}}^{\prime }}}=\sqrt{\frac{\mathrm{g}}{5}\mathrm{g}}=\sqrt{\frac{5}{4}}$

${\mathrm{T}}^{\prime }=\mathrm{T}\sqrt{\frac{5}{4}}=\frac{10}{2}\sqrt{5}$

${\mathrm{T}}^{\prime }=5\sqrt{5}$

Correct Answer is Option (A)

$\left|g_{eff}\right|=\left|\bar{g}-\bar{a}\right|$

$\Rightarrow g_{eff}=g\cos \theta$

$\Rightarrow T=2\pi \sqrt{\frac{l}{g_{eff}}}$

$=2\pi =\sqrt{\frac{L}{g\cos \theta }}$