JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 26)
512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V.
Respondre
128
Explicació
Let charge on each drop = q
radius = r
$$v = {{kq} \over r}$$
$$ \Rightarrow $$ $$2 = {{kq} \over r}$$
radius of bigger
$${4 \over 3}\pi {R^3} = 512 \times {4 \over 3}\pi {r^3}$$
$$R = 8r$$
$$ \therefore $$ $$v = {{k(512)q} \over R} = {{512} \over 8}{{kq} \over r} = {{512} \over 8} \times 2$$
$$ = 128V$$
radius = r
$$v = {{kq} \over r}$$
$$ \Rightarrow $$ $$2 = {{kq} \over r}$$
radius of bigger
$${4 \over 3}\pi {R^3} = 512 \times {4 \over 3}\pi {r^3}$$
$$R = 8r$$
$$ \therefore $$ $$v = {{k(512)q} \over R} = {{512} \over 8}{{kq} \over r} = {{512} \over 8} \times 2$$
$$ = 128V$$