Question
If Planck’s constant h, gravitational constant G, and
speed of light v are taken as fundamental units, the dimension of time is:Solution
Let the dimension of time [T] be expressed in terms of Planck's constant [h], gravitational constant [G], and speed of light [v] as: [T] = [h]a[G]b[v]c We need to find the values of a, b, and c. Let's write down the dimensions of h, G, and v in terms of mass [M], length [L], and time [T]: [h] (Planck's constant) has dimensions of energy × time, so: [h] = [M L2 T-2][T]=[M L2 T-1] [G] (Gravitational constant) is obtained from Newton's law of gravitation F = (Gm1m2)/r2 so G = (Fr2/​ m1m2): [G] = [M-1 L3 T-2] [v] (speed of light) has dimensions of length per time: [v] = [L T-1] Now, substitute these dimensions into the equation for [T]: [T] = [M L2 T-1]a [M-1 L3 T-2]b [L T-1]c [T] = [Ma L2a T-a] [M-b L3b T-2b] [Lc T-c] [T] = [Ma-b L2a+3b+c T-a-2b-c] For the dimensions to be equal on both sides, the powers of M, L, and T must be the same: For M: a − b = 0 ⟹ a = b (1) For L: 2a + 3b + c = 0 (2) For T: −a − 2b – c =1 (3) Substitute a = b from (1) into (2) and (3): From (2): 2a + 3a +c = 0 ⟹ 5a + c = 0 ⟹ c = −5a (4) From (3): −a−2a−c = 1 ⟹ −3a − c = 1 (5) Now substitute c = −5a from (4) into (5): −3a−(−5a)=1 −3a+5a=1 2a=1 a = 1/2 Since a = b, we have b=1/2​. Now substitute the value of a into equation (4) to find c: c = -5a = -5/2 The dimension of time is [h1/2 G1/2 v-5/2] Therefore, the correct answer is option (C).
2807, 1400, 697, 346, 171, 84, 41, ?
1, 6, 27, 124, ?, 3906, 27391
...5, 23, 18, 36, 31, ?
124, 128, ?, 135, 110, 146
12, 25, 51, ?, 207, 415Â
28, 43, 73, 118, ?, 253
37, 49, 63, ?, 97, 117
What will come in place of the question mark (?) in the following series?
24, 8, 15, ?, 12, 4
16.12 × 15.94 + 654.92 – 344.83 = ?× 5.95
What will come in place of the question mark (?) in the following series?
555, 430, ?, -320, -1320, -3320Â