Two masses 90 kg and 160 kg are separated by a distance of 5 m. The magnitude of intensity of the gravitational field at a point which is at a distance 3 m from the 90 kg mass and 4 m from the 160 kg mass is (Universal gravitational constant, .G=6.67 × 10-11 N - m 2 kg -2)

Question

Two masses 90 kg and 160 kg are separated by a distance of 5 m. The magnitude of intensity of the gravitational field at a point which is at a distance 3 m from the 90 kg mass and 4 m from the 160 kg mass is (Universal gravitational constant, .G=6.67 × 10-11 N - m 2 kg -2) (A) 94.3 × 10-10 N kg-1 (B) 9.43 × 10-10 N kg-1 (C) 9.43 × 10-12 N kg-1 (D) 94.3 × 10-12 N kg-1

Tardigrade Admin · Accepted Answer

A system of two masses is shown in the figure,

Here,

m_{1} = 90 k g

and

m_{2} = 160 k g

Gravitational field intensity due to mass

A

,

E_{A} = \frac{G M _{A}}{r _{C A}^{2}}

= G \frac{90}{( 3 ^{2} )} = 10 G \overset{r}{^}_{C A}

Similarly,

E_{B} = G \times \frac{160}{4 ^{2}} = 10 G \overset{r}{^}_{CB}

In

△ A BC

,

(A B)^{2} = (A C)^{2} + (BC)^{2}

(5)^{2} = (3)^{2} + (4)^{2}

Hence

Δ A BC

is a right angle triangle.
Hence, the resultant of

E_{A}

and

E_{B}

,

E = E_{A}^{2} + E_{B}^{2}

= (10 G)^{2} + (10 G)^{2} = 2 \times 10 G

Putting the value of

G

, we get

\Rightarrow E = 2 \times 10 \times 6.67 \times 1 0^{- 11}

= 9.43 \times 1 0^{- 10} N k g^{- 1}

$94.3 \times 1 0^{- 10} N k g^{- 1}$

$9.43 \times 1 0^{- 10} N k g^{- 1}$

$9.43 \times 1 0^{- 12} N k g^{- 1}$

$94.3 \times 1 0^{- 12} N k g^{- 1}$

Q. Two masses 90kg and 160kg are separated by a distance of 5m. The magnitude of intensity of the gravitational field at a point which is at a distance 3m from the 90kg mass and 4m from the 160kg mass is (Universal gravitational constant, G=6.67×10−11N−m2kg−2)

94.3×10−10Nkg−1

9.43×10−10Nkg−1

9.43×10−12Nkg−1

94.3×10−12Nkg−1

$94.3 \times 1 0^{- 10} N k g^{- 1}$

$9.43 \times 1 0^{- 10} N k g^{- 1}$

$9.43 \times 1 0^{- 12} N k g^{- 1}$

$94.3 \times 1 0^{- 12} N k g^{- 1}$