Dimension of \(\frac{1}{μ_0∈_0}\) should be equal to
- \(\frac{L}{T}\)
- \(\frac{T}{L}\)
- \(\frac{L^2}{T^2}\)
- \(\frac{T^2}{L^2}\)
The Correct Option is C
Solution and Explanation
Step 1: Use the relationship between µ0, ϵ0, and the speed of light.- Given $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, we know:
$\frac{1}{\mu_0 \epsilon_0} = c^2$.
Step 2: Analyze the dimensions.- Dimensional formula for c: [c] = $\frac{L}{T}$.- Hence, $[\frac{1}{\mu_0 \epsilon_0}]$ = $[c^2]$ = $\frac{L^2}{T^2}$.
Final Answer: The dimension is $\frac{L^2}{T^2}$
Top Questions on work, energy and power
- A body of mass 10 kg is moving with a speed of 4 m/s. It is brought to rest by a force in 5 seconds. Calculate the work done by the force.
- MHT CET - 2025
- Physics
- work, energy and power
- A force of 10 N is applied to move a body of mass 5 kg over a distance of 3 meters. Find the work done by the force.
- MHT CET - 2025
- Physics
- work, energy and power
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)):
- KCET - 2025
- Physics
- work, energy and power
- A 0.5 kg object is moving with a velocity of 10 m/s. What is its kinetic energy?
- MHT CET - 2025
- Physics
- work, energy and power
- A 50 kg person climbs a staircase of height 10 m. Calculate the work done by the person against gravity.
- MHT CET - 2025
- Physics
- work, energy and power
Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
- JEE Main - 2025
- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: