A radioactive sample ${S_1}$ having the activity ${A_1}$ has twice the number of nuclei as another sample ${S_2}$of activity ${A_2}$. If ${A_2} = 2{A_1}$, then the ratio of half-life of ${S_1}$ to the half-life of ${S_2}$ is

Activity, $A = \lambda N = \frac{{0.693}}{{{T_{1/2}}}}N$ Where 𝑇1/2 is the half-life of a radioactive sample, $\begin{array}{l} \therefore \,\,\,\,\frac{{{A_1}}}{{{A_2}}} = \frac{{{N_1}}}{{{T_1}}}\,\, \times \,\,\frac{{{T_2}}}{{{N_2}}}\\ \,\,\,\,\,\,\,\,\frac{{{T_1}}}{{{T_2}}} = \frac{{{A_2}}}{{{A_1}}}\,\, \times \,\,\frac{{{N_1}}}{{{N_2}}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2{A_1}}}{{{A_1}}}\,\, \times \,\,\frac{{2{N_2}}}{{{N_2}}} = \frac{4}{1} \end{array}$

The correct Option is "${\rm{4}}$"

The figure shows the displacement-time graph of a particle executing SHM. If the time period of oscillation is $2 \mathrm{~s}$ and amplitude is $10 \mathrm{~m}$, then equation of its SHM is

Here $A=10, \omega=\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi$ at $t=0 x=\frac{A}{2} \Rightarrow \phi=\frac{\pi}{6}$ So equation will be $x=10 \sin \left(\pi t+\frac{\pi}{6}\right)$

$x=10 \sin \left(\pi t+\frac{\pi}{3}\right)$

The correct Option is "$x=10 \sin \left(\pi t+\frac{\pi}{6}\right)$"

$x=10 \sin \left(2 \pi t+\frac{\pi}{6}\right)$

The K.E of a satellite is ${10^4}J$. It s P.E is

Potential energy of a satellite is the twice the kinetic energy with a negative $P.E = - 2K.E = - 2 \times {10^4}J$

The correct Option is "$ - 2 \times {10^4}\,J$"

The curves (a), (b) (c) and (d) show the variation between the applied potential difference $\left( V \right)$ and the photoelectric current $\left( i \right),$ at two different intensities of light $\left( {{I_1} > {I_2}} \right).$ In which figure is the correct variation shown

${I_1} > {I_2}$ (given) $ \Rightarrow {i_1} > {i_2}\;\left[ {\therefore i \propto I} \right]$ and stopping potential does not depend upon intensity. So its value will be same $\left( {{V_0}} \right)$

Two rods of same length and material transfer a given amount of heat in 12 s, when they are joined end to end (ie, in series). But when they are joined in parallel, they will transfer same heat under same conditions

$\Delta t = \frac{{\Delta Q\left( {\Delta x} \right)}}{{KA\left( {\Delta T} \right)}}$ When two rods of same length are joined in parallel, $A \to 2$ and $\left( {\Delta x} \right) \to \frac{1}{2}times$ $\therefore \,\,\Delta t\,\,becomes\frac{1}{4}times\,\,ie,\frac{1}{4} \times 12s = 3s$

The diagonals of a parallelogram are $\overrightarrow A = 2\hat i - 3\hat j + \hat k$ and $\overrightarrow B = 2\hat i + 4\hat j - \hat k$. Then the area of the parallelogram is

Given $\overrightarrow A = 2\hat i - 3\hat j + \hat k$ and $\overrightarrow B = 2\hat i + 4\hat j - \hat k$ The area of parallelogram $ = \frac{1}{2}\left| {{d_1} \times {d_2}} \right|$ $ = \frac{1}{2}\left| {\overrightarrow A \times \overrightarrow B } \right|$ .... (1) $\overrightarrow A \times \overrightarrow B = \left| {\begin{array}{*{20}{c}} i&j&k\\ 2&{ - 3}&1\\ { - 2}&4&{ - 1} \end{array}} \right| = i\left[ {\left( { - 3} \right)\left( { - 1} \right) - 1\left( 4 \right)} \right] - j\left[ {2\left( { - 1} \right) - 1\left( { - 2} \right)} \right] + k\left[ {2\left( 4 \right) - \left( { - 2} \right)\left( { - 3} \right)} \right]$ $\begin{array}{l} = i\left[ {3 - 4} \right] - j\left[ { - 2 + 2} \right] + k\left[ {8 - 6} \right]\\ = - \hat i + 0\hat j + 2\hat k\\ \left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt {1 + 4} = \sqrt 5 \end{array}$ From eq (1) area of parallelogram $ = \frac{{\sqrt 5 }}{2}$

The correct Option is "$\frac{{\sqrt 5 }}{2}$"

Which of the following statements is wrong?

If momentum is Zero ie, if p=0,then kinetic energy $K = \frac{{{p^2}}}{{2m}} = 0$ But potential energy cannot be zero, thus a body can have energy without momentum.

KE of a body is independent of the direction of motion

In an elastic collision of two bodies ,the momentum and energy of each body is conserved

The correct Option is "If two protons are brought towards each other the PE of the system decreases."

A body cannot have energy without momentum.

A horizontal force F is applied on a block mass m placed on a rough inclined plane of inclination $\theta $. The normal reaction N is

As is clear from figure $R = m{\rm{g}}\cos {\rm{\theta }} + F\sin {\rm{\theta }}$

$m{\rm{g}}\cos {\rm{\theta }} - F\cos {\rm{\theta }}$

The correct Option is "$m{\rm{g}}\cos {\rm{\theta }} + F\sin {\rm{\theta }}$"

Which of the following is wrongly matched?

Scientific principle of computer is digital circuit.

Optical fibres-Total internal reflection of light

Sonar-Reflection of ultrasonic waves

The correct Option is "Computers-Superconductivity"

The resistance of a conductor is $5\,\,ohm$ at $50\,{\,^0}{\rm{C}}$ and $6\,\,ohm$ at $100\,{\,^0}{\rm{C}}$. Its resistance at $0\,{\,^0}{\rm{C}}$ is

$\frac{{{R_1}}}{{{R_2}}} = \frac{{\left( {1\,\, + \,\,\alpha {t_1}} \right)}}{{\left( {1\,\, + \,\,\alpha {t_2}} \right)}} \Rightarrow \frac{5}{6} = \frac{{\left( {1\,\, + \,\,\alpha \,\, \times \,\,50} \right)}}{{\left( {1\,\, + \,\,\alpha \,\, \times \,\,100} \right)}} \Rightarrow \alpha = \frac{1}{{200}}\,\,{\rm{per}}{{\rm{ }}^0}{\rm{C}}$ Again by ${R_t} = {R_0}\left( {1 + \alpha t} \right)$ $ \Rightarrow 5 = {R_0}\left( {1 + \,\,\frac{1}{{200}}\,\, \times \,\,50} \right) \Rightarrow {R_0} = 4\,\,{\rm{\Omega }}$