- AWill remain unchanged.
- BWill be doubled.
- ✓Will be halved.
- DWill become four times the original.
Cut off wavelength is given by
$\lambda_{\text{min}}=\frac{\text{hc}}{\text{eV}'}$
where $h =$ Planck's constan
$c =$ speed of light
$e =$ charge on an electron
$V =$ potential difference applied to the tube
When potential difference $(V)$ applied to the tube is doubled, cutoff wavelength
$\big({\lambda}'_\text{min}\big)$ is given by
${\lambda}'_\text{min}=\frac{\text{hc}}{\text{e}(2\text{V})}$
$\Rightarrow\lambda'_\text{min}=\frac{\lambda_\text{min}}{2}$
Cuttoff wavelength does not depend on the separation between the filament and the target.
Thus, cutoff wavelength will be halved if the the potential difference applied to the tube is doubled.
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| Capacitor | Capacitance |
| $(A)$ Cylindrical capacitor | $(i)$ ${4\pi { \in _0}R}$ |
| $(B)$ Spherical capacitor | $(ii)$ $\frac{{KA{ \in _0}}}{d}$ |
| $(C)$ Parallel plate capacitor having dielectric between its plates | $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$ |
| $(D)$ Isolated spherical conductor | $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$ |