If we use permittivity varepsilon , resistance R, gravitational constant G and voltage V as fundamental physical quantities, then

Q: If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then

d $\epsilon^{-1}=L^{3} M T^{-2} Q^{-2}, R=L^{2} M T^{-1} Q^{-2}, V=M L^{2} T^{-2} Q^{-1}, G=L^{3} M^{-1} T^{-2}$ As the angular displacement ( $\theta$ ) is dimensionless so [displacement] $=\epsilon^{0} R^{0} G^{0} V^{0}$ Velocity, $[v]=L T^{-1}$ here $\epsilon^{-1} R^{-1} G^{0} V^{0}=L^{3} M T^{-2} Q^{-2} L^{-2} M^{-1} T^{1} Q^{2}=L T^{-1}=[v]$ dipole moment, $[p]=Q L$ here $\epsilon^{1} R^{0} G^{0} V^{1}=\left(L^{-3} M^{-1} T^{2} Q^{2}\right)\left(M L^{2} T^{-2} Q^{-1}\right)=Q L^{-1} \neq[p]$ force, $[f]=\epsilon^{-1} L^{-2} Q^{2}$ here $\epsilon^{1} R^{0} G^{0} V^{2}=\left(L^{-3} M^{-1} T^{2} Q^{2}\right)\left(M^{2} L^{4} T^{-4} Q^{-2}\right)=\left(L^{3} M T^{-2} Q^{-2}\right) L^{-2} Q^{2}=$ $\epsilon^{-1} L^{-2} Q^{2}=[f]$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then

A[angular displacement] $= \varepsilon^0R^0G^0V^0$
B[Velocity] = $\varepsilon ^{-1}R^{-1}G^0V^0$
C[force] = $ \varepsilon ^1R^0G^0V^2$
D
all of the above

Diffcult

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Download App

Similar Questions

1
The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right) \% .$ If the relative errors in measuring the mass and the diameter are $6.0 \%$ and $1.5 \%$ respectively, the value of $x$ is
View Solution
2
When the gap is closed without placing any object in the screw gauge whose least count is $0.005\ mm$, the $5^{th}$ division on its circular scale with the reference line on main scale, and when a small sphere is placed reading on main scale advances by $4$ divisions, whereas circular scale reading advances by five times to the corresponding reading when no object was placed. There are $200$ divisions on the circular scale. The radius of the sphere is .......... $mm$
View Solution
3
The dimensional formula of angular velocity is
View Solution
4
In the expression for time period $T$ of simple pendulum $T=2 \pi \sqrt{\frac{l}{g}}$, if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively then percentage error in acceleration due to gravity $g$ is equal to ......... $\%$
View Solution
5
In an experiment the angles are required to be measured using an instrument, $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half- a degree $(= 0.5^o )$, then the least count of the instrument is
View Solution
6
The dimensions of Planck's constant and angular momentum are respectively
View Solution
7
If the value of resistance is $10.845 \,ohm$ and the value of current is $3.23 \,amp$, the value of potential with significant numbers would be ........... $volt$
View Solution
8
Two resistors of resistances $R_1 = (300 \pm 3) \,\Omega $ and $R_2 = (500 \pm 4)$ are connected in series. The equivalent resistance of the series combination is
View Solution

Match List $I$ with List $II$ :

List $I$ (Physical Quantity)	List $II$ (Dimensional Formula)
$(A)$ Pressure gradient	$(I)$ $\left[ M ^0 L ^2 T ^{-2}\right]$
$(B)$ Energy density	$(II)$ $\left[ M ^1 L ^{-1} T ^{-2}\right]$
$(C)$ Electric Field	$(III)$ $\left[ M ^1 L ^{-2} T ^{-2}\right]$
$(D)$ Latent heat	$(IV)$ $\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right]$

Choose the correct answer from the options given below:

View Solution

10
The unit of angular acceleration in the SI system is
View Solution

Download our appand get started for free

Similar Questions

Download our app
and get started for free