JEE Advance - Physics (2020 - Paper 1 Offline - No. 18)
A uniform electric field, $$\overrightarrow E = - 400\sqrt 3 \widehat y$$ NC−1
is applied in a region. A charged particle of mass m
carrying positive charge q is projected in this region with an initial speed of 2$$\sqrt {10} $$ $$ \times $$ 106 ms−1
. This
particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown
schematically in the figure.
Take $${q \over m}$$ = 1010 Ckg−1 . Then
Take $${q \over m}$$ = 1010 Ckg−1 . Then
the particle will hit T if projected at an angle 45o
from the horizontal
the particle will hit T if projected either at an angle 30o or 60o
from the horizontal
time taken by the particle to hit T could be $$\sqrt {{5 \over 6}} $$ $$\mu $$s as well as $$\sqrt {{5 \over 2}} $$ $$\mu $$s
time taken by the particle to hit T is $$\sqrt {{5 \over 3}} $$ $$\mu $$s
Explanation
$$a = {{qE} \over m}{10^{10}} \times 400\sqrt 3 \,m/{s^2}$$
or $$a = 4\sqrt 3 \times {10^{12}}\,m/{s^2}$$
$$ \therefore $$ $$R = {{{u^2}\sin 2\theta } \over a}$$
$$ \Rightarrow \sin 2\theta = {{Ra} \over {{u^2}}}$$
or $$\sin 2\theta = {{5 \times 4\sqrt 3 \times {{10}^{12}}} \over {4 \times 10 \times {{10}^{12}}}}$$
$$ \Rightarrow \sin 2\theta = {{\sqrt 3 } \over 2}$$
$$ \Rightarrow 2\theta = 60^\circ ,\,120^\circ $$
$$ \Rightarrow \theta = 30^\circ $$ or $$\,60^\circ $$ for same range
$$ \therefore $$ $$T = {{2u\sin \theta } \over a}$$
$$ = {{2 \times 2\sqrt {10} \times {{10}^6}} \over {2 \times 4\sqrt 3 \times {{10}^{12}}}}$$
$$ \Rightarrow T = \sqrt {{{10} \over {12}}} \times {10^{ - 6}} \Rightarrow T = \sqrt {{5 \over 6}} \times {10^{ - 6}}$$
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