A) \[2.0\times {{10}^{-6}}m\]
B) \[1.7\times {{10}^{-6}}m\]
C) \[1.4\times {{10}^{-6}}m\]
D) \[1.1\times {{10}^{-6}}m\]
Correct Answer: B
Solution :
: For keeping the drop stationary, Weight of drop = Electrostatic force upwards Volume x density \[\times g=E\times 2e\] Or \[\left( \frac{4}{3}\pi {{R}^{3}} \right)\rho g=\frac{V\times 2e}{d}\] Or \[{{R}^{3}}=\frac{V\times 2e\times 3}{4\pi \rho gd}\] \[=\frac{(12000)\times (2\times 1.6\times {{10}^{-19}})\times 3}{4\times 3.14\times 900\times 10\times (2\times {{10}^{-2}})}\] Or \[{{R}^{3}}=\frac{16\times {{10}^{-18}}}{3.14}\] \[\Rightarrow \] \[{{R}^{3}}=5.09\times {{10}^{-18}}\] or \[R=1.7\times {{10}^{-6}}m\]
You need to login to perform this action.
You will be redirected in
3 sec
