Time period of a simple pendulum on earth's surface is ' T '. It time period becomes ' xT ' when taken to a height ' $2 R$ ' above earth's surface. The value of $x$ will be $(R=$ radius of earth)

The wire loop PQRSP formed by joining two semicircular wire of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre ' O ' is

A body of mass 1 kg begins to move under the action of a time dependent force $\overrightarrow{\mathrm{F}}=\left(\hat{\mathrm{t}}+2 t^2 \hat{\mathrm{j}}\right) \mathrm{N}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ axis. The power developed by above force at time $\mathrm{t}=3$ second will be

The period of oscillating simple pendulum is $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ where length ' $l$ ' is 100 cm with error 1 mm . Period is 2 second. The time of 100 oscillations is measured by a stopwatch of least count 0.1s. The percentage error in gravitational acceleration ' g ' is