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Dear Student, nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;Here #160;F=Gm1m2r2F ...

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Byju's Answer

Standard VI

Physics

Mass

Show #160;c...

Question

$S h o w c a l c u l a t i o n a n d m e t h o d . I f i n a s y s t e m, t h e f o r c e o f a t t r a c t i o n b e t w e e n t w o p o i n t m a s s e s o f 1 k g e a c h s i t u a t e d 1 k m a p a r t i s t a k e n a s a u n i t o f f o r c e a n d i s c a l l e d n o t w e n (n e w t o n w r i t t e n i n r e v e r s e o r d e r) & i f G = 6.67 \times 10^{- 11} N - m^{2} k g^{2} i n S I u n i t s t h e n w h i c h o f t h e f o l l o w i n g i s t r u e ? (A) 1 n o t w e n = 6.67 x 10^{- 11} n e w t o n (B) 1 n e w t o n = 6.67 x 10^{— 17} n o t w e n (C) 1 n o t w e n = 6.67 x 10^{- 17} n e w t o n (D) I n e w t o n = 6.67 x 10^{- 12} n o t w e n$

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Solution

Dear Student,
$H e r e F = \frac{G m_{1} m_{2}}{r^{2}} F = \frac{6.67 \times 10^{- 11} \times 1 \times 1}{1000^{2}} = 6.67 \times 10^{- 17} n e w t o n = 1 n o t w e n H e n c e 1 n o t w e n = 6.67 \times 10^{- 17} n e w t o n o p t i o n C$
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Similar questions

Q. If in a system, the force of attraction between two point masses of

1 k g

each situated

1 k m

apart is taken as a unit of force and is called notwen (Newton written in reverse order) and if

G = 6.67 \times 10^{- 11} N - m^{2} k g^{- 2}

in SI units then which of the following is true ?

Two spherical balls of mass 10 kg each are placed with their centers 10 cm apart. Find the gravitational force of attraction between them. Take $G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}$

Calculate the force between two masses of $50 k g$ and $100 k g$ separated by a distance of $10 m .$

Q. Prove that gravitational force of attraction between two boys (each weighing

35 k g

) sitting

1

metre apart is almost negligible as compared to the force of gravitation between either of the boys and the earth. Mass of the earth

= 6 \times 10^{24} k g

and radius of the earth

= 6.37 \times 10^{6} m

Take

G = 6.67 \times 10^{- 11} S I

units.

Calculate the force between two masses of 100 kg and 1000 kg separated by a distance of 10 m ( $G = 6.67 * 10^{- 11}$ N $\frac{m^{2}}{k g^{2}}$ ).

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Dear Student, Here F=Gm1m2r2F=6.67×10-11×1×110002=6.67×10-17newton=1notwenHence 1notwen=6.67×10-17newtonoption C Regards

Dear Student,
$H e r e F = \frac{G m_{1} m_{2}}{r^{2}} F = \frac{6.67 \times 10^{- 11} \times 1 \times 1}{1000^{2}} = 6.67 \times 10^{- 17} n e w t o n = 1 n o t w e n H e n c e 1 n o t w e n = 6.67 \times 10^{- 17} n e w t o n o p t i o n C$
Regards