STATEMENT-1
It is easier to pull a heavy object than to push it on a level ground.
STATEMENT-2
The magnitude of frictional force depends on the nature of the two surfaces in contact.

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
Both statement are independently true but 2 is not the correct explanation of 1.
The reason of statement 1 is that while pulling, a component of force reduces the normal reaction and hence reduces the friction force.

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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} .

S_{1} = \sum_{j = 1}^{10} j (j - 1)^{10} C_{j}, S_{2} = \sum_{j = 1}^{10} j .^{10} C_{j}, S_{3} = \sum_{j = 1}^{10} j^{2} .^{10} C_{j} .

S_{3} = 55 \times 2^{9} .

S_{1} = 90 \times 2^{5} a n d S_{2} = 10 \times 2^{8}

S_{1} = \sum_{j = 1}^{10} j (j - 1)^{10} C_{j}, S_{2} = \sum_{j = 1}^{10} j^{10} C_{j} and S_{3} = \sum_{j = 1}^{10} j^{2}^{10} C_{j} .

S_{3} = 55 \times 2^{9}

S_{1} = 90 \times 2^{8} and S_{2} = 10 \times 2^{8}

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