JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 6)
A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx2 , is given by :
$$Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) - BL} \right]$$
$$Gm\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$$
$$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) + BL} \right]$$
$$Gm\left[ {A\left( {{1 \over {a + L}} - {1 \over a}} \right) - BL} \right]$$
Explanation
_12th_January_Morning_Slot_en_6_1.png)
dm = (A + Bx2)dx
dF = $${{GMdm} \over {{x^2}}}$$
F = $$\int_a^{a + L} {{{GM} \over {{x^2}}}} $$ (A + Bx2)dx
= GM$$\left[ { - {A \over x} + Bx} \right]_a^{a + L}$$
= GM$$\left[ {A\left( {{1 \over a} - {1 \over {a + L}}} \right) + BL} \right]$$
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