Three point masses 9g, 8g and 4g are placed at the vertices A,B,C respectively of an isosceles triangle. What is the magnitude of gravitational field at a point which is the foot of the perpendicular drawn from vertex A on to BC? (AB = AC = 5cm, BC = 8cm).

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$4 \times 10^{- 11} d y n e / g$

$6.88 \times 10^{- 8} d y n e / g$

$9 \times 10^{- 5} N / k g$

$6.67 \times 10^{- 11} d y n e / k g$

answer is B.

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Detailed Solution

Value of G in C.G.S = $6.67 \times 10^{- 8} d y n e c m^{2} / g^{2}$
Field at point P is vector sum of fields due to all individual masses.
${\vec{E}}_{P} = {\vec{E}}_{A} + {\vec{E}}_{B} + {\vec{E}}_{C}$
$E_{A} = \frac{G \times 9}{{(3)}^{2}} = \frac{6.67 \times 10^{- 8} \times 9}{9} = 6.67 \times 10^{- 8} d y n e / g$
$E_{B} = \frac{G \times 8}{{(4)}^{2}} = \frac{6.67 \times 10^{- 8} \times 8}{16} = \frac{6.67}{2} \times 10^{- 8} d y n e / g$
$E_{C} = \frac{G \times 4}{4^{2}} = \frac{6.67 \times 10^{- 8} \times 4}{16} = \frac{6.67}{4} \times 10^{- 8} d y n e / g$
$= 1.667 \times 10^{- 8} d y n e / g$
${\vec{E}}_{B}$ and ${\vec{E}}_{C}$ act on the same line in opp. Direction since $E_{B} > E_{C}$
The result of ${\vec{E}}_{B} + {\vec{E}}_{C}$ will be towards B and will have a magnitude $E_{B} - E_{C}$
$E_{B} - E_{C} = 1.668 \times 10^{- 8} d y n e / g$
The net field at point P is
$E = \sqrt{E_{A}^{2} + {(E_{B} - E_{C})}^{2}}$
$N e t g r a v i t a t i o n a l f i e l d a t P = E = \sqrt{{(6.67 \times 10^{- 8})}^{2} + {(1.667 \times 10^{- 8})}^{2}} = 6.88 \times 10^{- 8} d y n e / g$