Assertion :Gravitational potential is maximum at infinity. Reason: Gravitational potential is the amount of work done to shift a unit mass from infinity to a given point in gravitational attraction force field.

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is False

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STATEMENT-1 is False, STATEMENT-2 is True

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Solution

The correct option is B STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
Gravitational potential at inifinity $G_{i} = 0$ .
Gravitational potential at r is $G_{r} = \frac{- G M m}{r} < 0 \to G_{r} < G_{i}$ .
Work done = change in gravitational potential energy.
Thus , both the statements are true but $2$ does not imply $1$ .

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

\sim (p \leftrightarrow\sim q)

p \leftrightarrow q

\sim (p \leftrightarrow\sim q)

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