In this problem, we need to find the electrostatic potential at the center, at a distance of 5 cm from the center, and at a distance of 15 cm from the center of a uniformly charged spherical shell with a given radius \( R = 10 \, \text{cm} \). The surface potential is 120 V.
The concepts involved in this problem include:
\(V = \frac{kQ}{r}\)
where \( k \) is Coulomb's constant and \( Q \) is the total charge on the shell.
Now, let's calculate the potential at each region mentioned:
\(V = \frac{kQ}{15 \, \text{cm}}\)
where \( V = 120 \, \text{V} \, \text{at} \, r = 10 \, \text{cm} \). Therefore, the potential at 15 cm is less than the surface potential and is calculated by:
\(V_{\text{15 cm}} = \left(\frac{R}{15 \, \text{cm}}\right) \times 120 = \left(\frac{10}{15}\right) \times 120 = 80 \, \text{V}\)
Thus, the potentials are:
The correct option is 120V, 120V, 80V.
To solve this question, we need to determine the electrostatic potential at three different positions relative to a uniformly charged spherical shell. These positions are:
The given information includes:
Now, let's analyze the potential at each point:
Therefore, the potentials at the distinct positions are:
Hence, the correct answer is: 120V, 120V, 80V.



Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are: