JEE MAIN - Chemistry (2021 - 26th February Evening Shift - No. 24)
A ball weighing 10 g is moving with a velocity of 90 ms$$-$$1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is ___________ $$\times$$ 10$$-$$33 m. (Rounded off to the nearest integer)
[Given : h = 6.63 $$\times$$ 10$$-$$34 Js]
[Given : h = 6.63 $$\times$$ 10$$-$$34 Js]
Answer
1
Explanation
m = 10 g = 10$$-$$2 Kg
v = 90 m/sec.
$$\Delta$$ = v $$\times$$ 5% = 90 $$\times$$ $${5 \over {100}}$$ = 4.5 m/sec
m . $$\Delta$$v . $$\Delta$$x $$ \ge $$ $${h \over {4\pi }}$$
10$$-$$2 $$\times$$ 4.5 $$\times$$ $$\Delta$$x $$ \ge $$ $${{6.63 \times 3 \times {{10}^{ - 34}}} \over {4 \times {{22} \over 7}}}$$
$$\Delta$$x $$ \ge $$ $${{6.63 \times 7 \times 2 \times {{10}^{ - 34}}} \over {9 \times 4 \times 22 \times {{10}^{ - 2}}}}$$
$$\Delta$$x $$ \ge $$ 1.17 $$\times$$ 10$$-$$33 = x $$\times$$ 10$$-$$33
x = 1.17 $$ \simeq $$ 1
v = 90 m/sec.
$$\Delta$$ = v $$\times$$ 5% = 90 $$\times$$ $${5 \over {100}}$$ = 4.5 m/sec
m . $$\Delta$$v . $$\Delta$$x $$ \ge $$ $${h \over {4\pi }}$$
10$$-$$2 $$\times$$ 4.5 $$\times$$ $$\Delta$$x $$ \ge $$ $${{6.63 \times 3 \times {{10}^{ - 34}}} \over {4 \times {{22} \over 7}}}$$
$$\Delta$$x $$ \ge $$ $${{6.63 \times 7 \times 2 \times {{10}^{ - 34}}} \over {9 \times 4 \times 22 \times {{10}^{ - 2}}}}$$
$$\Delta$$x $$ \ge $$ 1.17 $$\times$$ 10$$-$$33 = x $$\times$$ 10$$-$$33
x = 1.17 $$ \simeq $$ 1
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