Question

e
(c)roT
(d)r301)
NUMERICALS :-
(4 X 5)
Q.1) Force of attraction between two masses is 100N. Find the new force when
Distance b/w the masses is doubled
ii) each mass is doubled and
each mass is doubled and the distance is halved​

Answer 1

Answer:

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$