Question

❤❤❤❤Answer this question ❤❤❤​

Answer 1

Answer:

dimention of length (l) - [M⁰L¹T⁰]

dimention of

acceleration due

gravity (g) - [M⁰L¹T⁻²]

now, dimention of,

$l \sqrt{ \frac{l}{g} }$

=[M⁰L¹T⁰] ([M⁰L¹T⁰]/[M⁰L¹T⁻²])½

=[M⁰L¹T⁰] [M⁰L½T⁰]/[M⁰L½T⁻¹]

=[M⁰L¹⁺½T⁰]/ [M⁰L½T⁻¹]

=[M⁰L^3/2T⁰] [M⁰L⁻½T¹]

=[M⁰L¹T¹]

Hence the dimension is [M⁰L¹T¹]

Answer 2

Answer:

hii

kaisa h tu

bhul gya kya merko.........