JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 22)
Time (T), velocity (C) and angular momentum (h) are chosen as fundamentalquantities instead of mass, length and time. In terms of these, the dimensions of mass would be :
[M] = [T$$-$$1 C$$-$$2 h]
[M] = [T$$-$$1 C2 h]
[M] = [T$$-$$1 C$$-$$2 h$$-$$1]
[M] = [T C$$-$$2 h]
Explanation
Let,
M $$ \propto $$ Tx Cy hz
$$\therefore\,\,\,$$ [M1LoTo] = [T1]x [L1 T$$-$$1]y [M1L2T$$-$$1]z
[M1Lo To] = [Mz Ly + 2z Tx$$-$$y$$-$$z]
By comparing both sides we get,
z = 1
y + 2z = 0
x $$-$$ y $$-$$ z = 0
$$\therefore\,\,\,$$ y = $$-$$ 2z = $$-$$ 2
x = y + z = $$-$$2 + 1 = $$-$$1
$$\therefore\,\,\,$$ [M] = [M$$-$$1 C$$-$$2 h1]
M $$ \propto $$ Tx Cy hz
$$\therefore\,\,\,$$ [M1LoTo] = [T1]x [L1 T$$-$$1]y [M1L2T$$-$$1]z
[M1Lo To] = [Mz Ly + 2z Tx$$-$$y$$-$$z]
By comparing both sides we get,
z = 1
y + 2z = 0
x $$-$$ y $$-$$ z = 0
$$\therefore\,\,\,$$ y = $$-$$ 2z = $$-$$ 2
x = y + z = $$-$$2 + 1 = $$-$$1
$$\therefore\,\,\,$$ [M] = [M$$-$$1 C$$-$$2 h1]
Comments (0)
