If e and g respectively denote energy and gravitational constant

The correct option is D angle [ E ] = [ M L 2 T − 2 ] , [ m ] = [ M ] [ l ] = [ M L 2 T − 1 ] , [ G ] = [ M − 1 L 3 T − 2 ] ∴ ( E l 2 m 5 G 2 ) = [ M L 2 T − 2 ] [ M 2 L 4 T − 2 ] [ M 5 ] [ M − 2 L 6 T − 4 ] = [ M ∘ L ∘ T ∘ ] As angle has no dimensions, therefore E l 2 m 5 G 2 has the same dimensions as that of angle.

If E and G, respectively, denote energy and gravitational constant, thenEGhas the dimensions of: 1.ML0T0 2.M2L-2T-1 3.M2L-1T0 4.ML-1T-1 Units and Measurement Physics NEET Practice Questions, MCQs, Past Year Questions (PYQs), NCERT Questions, Question Bank, Class 11 and Class 12 Questions, and PDF solved with answers Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length ( l ), the mass of the bob ( m ) and acceleration due to gravity ( g ). Expression for its time period is: 1. T = 1 2 π l g 2. T = 2 π l g 3. T = 2 π l g 4. T = 2 π g l Let us consider an equation \(\fracmv^2=mgh\)where \(m\) is themass of the body, \(v\) its velocity, \(g\) is the acceleration due to gravity and \(h\) is the height. The equation is: 1. dimensionally correct. 2. dimensionally incorrect. 3. can not be checked by dimensional analysis. 4. can't say anything. The SI unit of energy is J = kgm 2 s - 2 ; that of speed v is ms - 1 and of acceleration a is ms - 2 . Which of the formula for kinetic energy ( K ) given below can you rule out on the basis of dimensional arguments ( m stands for the mass of the body)? (a) K = m 2 v 3 (b) K = 1 / 2 mv 2 (c) K = ma (d) K = (3/16) mv 2 (e) K = (1/2) mv 2 +ma 1. (a), (c) & (d) 2. (b) & (d) 3.(a), (c), (d) & (e) 4.(a), (c) & (e) The relative error in \(Z,\)if \(Z=\frac\)

`"Given, expression is "P=EL^(2)m^(-5)G^(-2)` `"where E is energy "[E]=[ML^(2)T^(-2)]` `"m is mass "[m]=[M]` `"L is angular momentum "[L]=[ML^(2)T^(-1)]` `"G is gravitational constant "[G]=[M^(-1)L^(3)T^(-2)]` Substituting dimensions of each term in the given expression, `[P]=[ML^(2)T^(-2)]xx[ML^(2)T^(_1)]^(2)xx[M]^(-5)xx[M^(-1)L^(3)T^(-2)]^(-2)` `=[M^(1+2-5+2)L^(2+4-6)T^(-2-2+4)]=[M^(0)L^(0)T^(0)]` Therefore, P is a dimensionless quantity.

Given quantity is E J 2 M 5 G 2 .......(i) where dimensions of the various given quantities are Dimensions of E = [ M L 2 T − 2 ] Dimensions of J = [ M L 2 T − 1 ] Dimension of M = [ M ] Dimension of G = [ M − 1 L 3 T − 2 ] Now, on putting these dimensions in Eq. (i), we have = [ M L 2 T − 2 ] [ M L 2 T − 1 ] 2 [ M 5 ] [ M − 1 L 3 T − 2 ] 2 = [ M 3 L 6 T − 2 ] [ M 3 L 6 T − 2 ] = dimensionless Since, angle is a dimensionless quantity

`[E]=[ML^(2)T^(-2)]` `[M]=[M],[L]=[ML^(2)T^(-1)]` ltbr. `[G]=[M^(-1)L^(3)T^(-2)]` `[(E^(2)L^(2)]/(M^(5)G^(2))]=([ML^(2)T^(-2)]^(2)[ML.^(2)T^(-1)]^(2))/([M]^(5)[M^(-1)L^(3)T^(-2)]^(2))` `=([M^(2)L^(4)T^(-4)][M^(2)L^(4)T^(-2)])/([M^(5)][M^(-2)L^(6)T^(-4)])=([M^(4)L^(8)T^(-6)])/([M^(3)L^(6)T^(-4)])=[ML^(2)T^(-2)]` The quantity `((E^(2)L^(2))/(M^(5)G^(2)))` hs the dimensions of energy. None of the given option is correct. If the quantity `(EL^(2))/(M^(5)G^(2)))` is given in place of quantity `((E^(2)L^(2))/(M^(5)G^(2)))` , then the option (1) is correct. `[(EL^(2))/(M^(5)G^(2))]=([ML^(2)T^(-2)][ML^(2)T^(-1)]^(2))/([M^(5)][M^(-1)L^(3)T^(-2)]^(2))` `=([ML^(2)T^(-2)][M^(2)L^(4)T^(-2)])/([M^(5)[M^(-2)L^(6)T^(-4)])]=([M^(3)L^(6)T^(-4)])/([M^(3)L^(6)T^(-4)])` `[M^(0)L^(0)T^(0)]= "Angle"`