- $\text{zero}$
- $\frac{1}{2}\text{Bl}\omega^2$
- $\text{Bl}\omega ^2$
- $2\text{B}\text{l}\omega^2$
Explanation:

Emf at both end is same $=\frac{1}{8}\text{Bwl}^2$
So the potential difference between the two ends of therod is zero.
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A highly conducting ring of radius R is perpendicular to and concentric with the axis of a long solenoid as shown in fig. The ring has a narrow gap of width d in its circumference. The solenoid has cross sectional area A and a uniform internal field of magnitude B0. Now beginning at t = 0, the solenoid current is steadily increased to so that the field magnitude at any time t is given by B(t) = B0 + at where α > 0. Assuming that no charge can flow across the gap, the end of ring which has excess of positive charge and the magnitude of induced e.m.f. in the ring are respectively

|
(a) X, A𝛼 |
(b) X, pR2𝛼 |
(c) Y, 𝜋A2𝛼 |
(d) Y, pR2𝛼 |
An electron (charge = 1.6
) is accelerated through a potential of 100,000 V. The energy acquired by the electron is
|
(a) 1.6 |
(b) 1.6 |
(c) 0.53 |
(d) 1.6 |
In vacuum an electron of energy 10 keV hits tungsten target, then emitted radiation will be
|
(a) Cathode rays |
(b) X-rays |
(c) Infrared rays |
(d) Visible spectrum |

Biaxial crystal among the following is
|
(a) Calcite |
(b) Quartz |
(c) Selenite |
(d) Tourmaline |
A certain charge liberates 0.8 gm of
. The same charge will liberate how many gm of silver
|
(a) 108 gm |
(b) 10.8 gm |
(c) 0.8 gm |
(d) |