The velocity of a freely falling body changes as g-ph-q where g is acceleration due to gravity and h is the height- The values of p and q are

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Toppr Admin · Accepted Answer

Correct option is B- -dfrac-1-2-dfrac-1-2- v -propto g-ph-q- -given-By substituting the dimension of each quantity and comparing the powers in both sides-160-we get -LT-1-LT-2-p-L-q-Rightarrow p-q-1-2p-1- -therefore p-dfrac-1-2-q-dfrac-1-2-

The velocity of a freely falling body changes as $$g^ph^q$$ where g is acceleration due to gravity and h is the height. The values of p and q are

Correct option is B. $$\dfrac{1}{2},\dfrac{1}{2}$$
$$ v \propto g^ph^q$$ (given)

By substituting the dimension of each quantity and comparing the powers in both sides

we get $$[LT^{-1}]=[LT^{-2}]^p[L]^q$$

$$\Rightarrow p+q=1,-2p=-1$$

$$ \therefore p=\dfrac{1}{2},q=\dfrac{1}{2}$$

The velocity of a freely falling body changes as $$g^ph^q$$ where g is acceleration due to gravity and h is the height. The values of p and q are

Correct option is B. $$\dfrac{1}{2},\dfrac{1}{2}$$$$ v \propto g^ph^q$$ (given)By substituting the dimension of each quantity and comparing the powers in both sides we get $$[LT^{-1}]=[LT^{-2}]^p[L]^q$$$$\Rightarrow p+q=1,-2p=-1$$$$ \therefore p=\dfrac{1}{2},q=\dfrac{1}{2}$$

Correct option is B. $$\dfrac{1}{2},\dfrac{1}{2}$$
$$ v \propto g^ph^q$$ (given)

By substituting the dimension of each quantity and comparing the powers in both sides

we get $$[LT^{-1}]=[LT^{-2}]^p[L]^q$$

$$\Rightarrow p+q=1,-2p=-1$$

$$ \therefore p=\dfrac{1}{2},q=\dfrac{1}{2}$$