
- A

- B

- ✓

- D







$A=\frac{1}{2} \frac{(\text { stress })^{2}}{Y}=$ constant
$\frac{_{1}}{2} \frac{\left(\frac{M}{L} \frac{x}{A} g\right)^{2}}{Y}=$ constant
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[Use $hc =1242 eV nm ; 1 nm =10^{-9} m , h$ and $c$ are Planck's constant and speed of light, respectively]
$(1)$ $\lambda_0=418 nm$
$(2)$ The ratio of kinetic energy of the electron in the state $n = m$ to the state $n =1$ is $\frac{1}{4}$
$(3)$ $m =2$
$(4)$ $\Delta p_{ a } / \Delta p _{ o }=\frac{1}{2}$
Assertion $A$: The potential ( $V$ ) at any axial point, at $2 \mathrm{~m}$ distance ( $r$ ) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6} \mathrm{C} \mathrm{m}$, is $\pm 9 \times 10^3 \mathrm{~V}$.
(Take $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Sl}$ units)
Reason $R$: $V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}$, where $r$ is the distance of any axial point, situated at $2 \mathrm{~m}$ from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below: