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PHXI02:UNITS AND MEASUREMENTS

367276 If the velocity of light $c$ , universal gravitational constant $G$ and Planck's constant $h$ are chosen as fundamental quantities. The dimensions of mass in the new system is

1

[h^{1} c^{1} G^{- 1}]

2

[h^{- 1 / 2} c^{1 / 2} G^{1 / 2}]

3

[h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]

4

[h^{1 / 2} c^{- 1 / 2} G^{1}]

Explanation:

As per question
$G = [M^{- 1} L^{3} T^{- 2}],$
$h = [M L^{2} T^{- 1}], c = [L T^{- 1}]$
Mass $\propto h^{a} G^{b} c^{c}$
$[M^{1} L^{0} T^{0}] = {[M L^{2} T^{- 1}]}^{a}$
${[M^{- 1} L^{3} T^{- 2}]}^{b} {[L T^{- 1}]}^{c}$
$[M^{1} L^{0} T^{0}] = [M^{a - b} L^{2 a + 3 b + c} T^{- a - 2 b - c}]$
$a - b = 1 (1)$
$2 a + 3 b + c = 0 (2)$
$- a - 2 b - c = 0 (3)$
By solving, $a = \frac{1}{2}, b = \frac{- 1}{2}, c = \frac{1}{2}$
$∴ m = [h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]$
So, correct option is (3).

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367277 Time $(T),$ velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be

1

[M] = [T C^{- 2} h]

2

[M] = [T^{- 1} C^{- 2} h]

3

[M] = [T^{- 1} C^{2} h]

4

[M] = [T^{- 1} C^{- 2} h^{- 1}]

Explanation:

Let mass, related as $M \propto T^{x} C^{y} h^{z}$
$M^{1} L^{0} T^{0} = {(T^{'})}^{x} {(L^{1} T^{- 1})}^{y} {(M^{1} L^{2} T^{- 1})}^{z}$
$M^{1} L^{0} T^{0} = M^{z} L^{y + 2 z} \times T^{x - y - z}$
$z = 1$
$y + 2 z = 0 x - y - z = 0$
$y = - 2 x + 2 - 1 = 0$
$x = - 1$
$M = [T^{- 1} C^{- 2} h^{1}]$

JEE - 2017

PHXI02:UNITS AND MEASUREMENTS

367278 If $y = a \sin θ$ , then

1 the unit of

y

is same as

a

2 the unit of

θ

is steradian

3

θ

is unitless

4 None of the above

Explanation:

$\sin θ$ is a ratio of lengths.
$∴$ Unit of $y =$ Unit of $a$ .
The SI unit of angle is radian.

PHXI02:UNITS AND MEASUREMENTS

367279 Given that force $(F)$ is given by $F = P t^{- 1} + Q t$ . Here $t$ is time. The unit of $P$ is same as that of

1 Acceleration

2 Momentum

3 Velocity

4 Displacement

Explanation:

$F = P t^{- 1} + Q t$
$[F] = [{P t}^{- 1}] = [Q t]$
$\Rightarrow [P] = [F] [t] = [M L T^{- 2}] [t]$
$= [M L T^{- 1}] = [p]$
The units of $P$ is same as that of momentum.
So correct option is (2)

PHXI02:UNITS AND MEASUREMENTS

367276 If the velocity of light $c$ , universal gravitational constant $G$ and Planck's constant $h$ are chosen as fundamental quantities. The dimensions of mass in the new system is

1

[h^{1} c^{1} G^{- 1}]

2

[h^{- 1 / 2} c^{1 / 2} G^{1 / 2}]

3

[h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]

4

[h^{1 / 2} c^{- 1 / 2} G^{1}]

Explanation:

As per question
$G = [M^{- 1} L^{3} T^{- 2}],$
$h = [M L^{2} T^{- 1}], c = [L T^{- 1}]$
Mass $\propto h^{a} G^{b} c^{c}$
$[M^{1} L^{0} T^{0}] = {[M L^{2} T^{- 1}]}^{a}$
${[M^{- 1} L^{3} T^{- 2}]}^{b} {[L T^{- 1}]}^{c}$
$[M^{1} L^{0} T^{0}] = [M^{a - b} L^{2 a + 3 b + c} T^{- a - 2 b - c}]$
$a - b = 1 (1)$
$2 a + 3 b + c = 0 (2)$
$- a - 2 b - c = 0 (3)$
By solving, $a = \frac{1}{2}, b = \frac{- 1}{2}, c = \frac{1}{2}$
$∴ m = [h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]$
So, correct option is (3).

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367277 Time $(T),$ velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be

1

[M] = [T C^{- 2} h]

2

[M] = [T^{- 1} C^{- 2} h]

3

[M] = [T^{- 1} C^{2} h]

4

[M] = [T^{- 1} C^{- 2} h^{- 1}]

Explanation:

Let mass, related as $M \propto T^{x} C^{y} h^{z}$
$M^{1} L^{0} T^{0} = {(T^{'})}^{x} {(L^{1} T^{- 1})}^{y} {(M^{1} L^{2} T^{- 1})}^{z}$
$M^{1} L^{0} T^{0} = M^{z} L^{y + 2 z} \times T^{x - y - z}$
$z = 1$
$y + 2 z = 0 x - y - z = 0$
$y = - 2 x + 2 - 1 = 0$
$x = - 1$
$M = [T^{- 1} C^{- 2} h^{1}]$

JEE - 2017

PHXI02:UNITS AND MEASUREMENTS

367278 If $y = a \sin θ$ , then

1 the unit of

y

is same as

a

2 the unit of

θ

is steradian

3

θ

is unitless

4 None of the above

Explanation:

$\sin θ$ is a ratio of lengths.
$∴$ Unit of $y =$ Unit of $a$ .
The SI unit of angle is radian.

PHXI02:UNITS AND MEASUREMENTS

367279 Given that force $(F)$ is given by $F = P t^{- 1} + Q t$ . Here $t$ is time. The unit of $P$ is same as that of

1 Acceleration

2 Momentum

3 Velocity

4 Displacement

Explanation:

$F = P t^{- 1} + Q t$
$[F] = [{P t}^{- 1}] = [Q t]$
$\Rightarrow [P] = [F] [t] = [M L T^{- 2}] [t]$
$= [M L T^{- 1}] = [p]$
The units of $P$ is same as that of momentum.
So correct option is (2)

PHXI02:UNITS AND MEASUREMENTS

367276 If the velocity of light $c$ , universal gravitational constant $G$ and Planck's constant $h$ are chosen as fundamental quantities. The dimensions of mass in the new system is

1

[h^{1} c^{1} G^{- 1}]

2

[h^{- 1 / 2} c^{1 / 2} G^{1 / 2}]

3

[h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]

4

[h^{1 / 2} c^{- 1 / 2} G^{1}]

Explanation:

As per question
$G = [M^{- 1} L^{3} T^{- 2}],$
$h = [M L^{2} T^{- 1}], c = [L T^{- 1}]$
Mass $\propto h^{a} G^{b} c^{c}$
$[M^{1} L^{0} T^{0}] = {[M L^{2} T^{- 1}]}^{a}$
${[M^{- 1} L^{3} T^{- 2}]}^{b} {[L T^{- 1}]}^{c}$
$[M^{1} L^{0} T^{0}] = [M^{a - b} L^{2 a + 3 b + c} T^{- a - 2 b - c}]$
$a - b = 1 (1)$
$2 a + 3 b + c = 0 (2)$
$- a - 2 b - c = 0 (3)$
By solving, $a = \frac{1}{2}, b = \frac{- 1}{2}, c = \frac{1}{2}$
$∴ m = [h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]$
So, correct option is (3).

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367277 Time $(T),$ velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be

1

[M] = [T C^{- 2} h]

2

[M] = [T^{- 1} C^{- 2} h]

3

[M] = [T^{- 1} C^{2} h]

4

[M] = [T^{- 1} C^{- 2} h^{- 1}]

Explanation:

Let mass, related as $M \propto T^{x} C^{y} h^{z}$
$M^{1} L^{0} T^{0} = {(T^{'})}^{x} {(L^{1} T^{- 1})}^{y} {(M^{1} L^{2} T^{- 1})}^{z}$
$M^{1} L^{0} T^{0} = M^{z} L^{y + 2 z} \times T^{x - y - z}$
$z = 1$
$y + 2 z = 0 x - y - z = 0$
$y = - 2 x + 2 - 1 = 0$
$x = - 1$
$M = [T^{- 1} C^{- 2} h^{1}]$

JEE - 2017

PHXI02:UNITS AND MEASUREMENTS

367278 If $y = a \sin θ$ , then

1 the unit of

y

is same as

a

2 the unit of

θ

is steradian

3

θ

is unitless

4 None of the above

Explanation:

$\sin θ$ is a ratio of lengths.
$∴$ Unit of $y =$ Unit of $a$ .
The SI unit of angle is radian.

PHXI02:UNITS AND MEASUREMENTS

367279 Given that force $(F)$ is given by $F = P t^{- 1} + Q t$ . Here $t$ is time. The unit of $P$ is same as that of

1 Acceleration

2 Momentum

3 Velocity

4 Displacement

Explanation:

$F = P t^{- 1} + Q t$
$[F] = [{P t}^{- 1}] = [Q t]$
$\Rightarrow [P] = [F] [t] = [M L T^{- 2}] [t]$
$= [M L T^{- 1}] = [p]$
The units of $P$ is same as that of momentum.
So correct option is (2)

PHXI02:UNITS AND MEASUREMENTS

367276 If the velocity of light $c$ , universal gravitational constant $G$ and Planck's constant $h$ are chosen as fundamental quantities. The dimensions of mass in the new system is

1

[h^{1} c^{1} G^{- 1}]

2

[h^{- 1 / 2} c^{1 / 2} G^{1 / 2}]

3

[h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]

4

[h^{1 / 2} c^{- 1 / 2} G^{1}]

Explanation:

As per question
$G = [M^{- 1} L^{3} T^{- 2}],$
$h = [M L^{2} T^{- 1}], c = [L T^{- 1}]$
Mass $\propto h^{a} G^{b} c^{c}$
$[M^{1} L^{0} T^{0}] = {[M L^{2} T^{- 1}]}^{a}$
${[M^{- 1} L^{3} T^{- 2}]}^{b} {[L T^{- 1}]}^{c}$
$[M^{1} L^{0} T^{0}] = [M^{a - b} L^{2 a + 3 b + c} T^{- a - 2 b - c}]$
$a - b = 1 (1)$
$2 a + 3 b + c = 0 (2)$
$- a - 2 b - c = 0 (3)$
By solving, $a = \frac{1}{2}, b = \frac{- 1}{2}, c = \frac{1}{2}$
$∴ m = [h^{1 / 2} c^{1 / 2} G^{- 1 / 2}]$
So, correct option is (3).

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367277 Time $(T),$ velocity $(C)$ and angular momentum $(h)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be

1

[M] = [T C^{- 2} h]

2

[M] = [T^{- 1} C^{- 2} h]

3

[M] = [T^{- 1} C^{2} h]

4

[M] = [T^{- 1} C^{- 2} h^{- 1}]

Explanation:

Let mass, related as $M \propto T^{x} C^{y} h^{z}$
$M^{1} L^{0} T^{0} = {(T^{'})}^{x} {(L^{1} T^{- 1})}^{y} {(M^{1} L^{2} T^{- 1})}^{z}$
$M^{1} L^{0} T^{0} = M^{z} L^{y + 2 z} \times T^{x - y - z}$
$z = 1$
$y + 2 z = 0 x - y - z = 0$
$y = - 2 x + 2 - 1 = 0$
$x = - 1$
$M = [T^{- 1} C^{- 2} h^{1}]$

JEE - 2017

PHXI02:UNITS AND MEASUREMENTS

367278 If $y = a \sin θ$ , then

1 the unit of

y

is same as

a

2 the unit of

θ

is steradian

3

θ

is unitless

4 None of the above

Explanation:

$\sin θ$ is a ratio of lengths.
$∴$ Unit of $y =$ Unit of $a$ .
The SI unit of angle is radian.

PHXI02:UNITS AND MEASUREMENTS

367279 Given that force $(F)$ is given by $F = P t^{- 1} + Q t$ . Here $t$ is time. The unit of $P$ is same as that of

1 Acceleration

2 Momentum

3 Velocity

4 Displacement

Explanation:

$F = P t^{- 1} + Q t$
$[F] = [{P t}^{- 1}] = [Q t]$
$\Rightarrow [P] = [F] [t] = [M L T^{- 2}] [t]$
$= [M L T^{- 1}] = [p]$
The units of $P$ is same as that of momentum.
So correct option is (2)

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