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Chapter 1: Problem 130

The dimensional formula of plank's constant is ........... (a) \(\mathrm{M}^{3} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3}\)

Short Answer

Expert verified
The short answer is: The dimensional formula of Planck's constant is \(\mathrm{M^1L^2T^{-1}}\).

Step by step solution

01

Recall the formula for energy

In problems like this, when dealing with dimensions, it is common to use the formula for kinetic energy (E) as a reference. The formula for kinetic energy is given by: \[E = \frac{1}{2}mv^{2}\] Where \(m\) is the mass (M) with dimensions \(\mathrm{M^1}\), and \(v\) is the velocity with dimensions \(\mathrm{LT^{-1}}\).
02

Find the dimensional formula for energy

Substitute mass and velocity dimensions in the kinetic energy formula: \[\mathrm{E} = \mathrm{M^1} \cdot (\mathrm{LT^{-1}})^2\] Simplify to get the dimensions for energy: \[\mathrm{E} = \mathrm{M^1L^2T^{-2}}\]
03

Multiply the dimensional formula for energy with time

Planck's constant has dimensions of energy multiplied by time. So, multiply the dimensions of energy by the dimensions of time (\(\mathrm{T}\)): \[\mathrm{h} = \mathrm{M^1L^2T^{-2}} \cdot \mathrm{T}\] Simplify to get the dimensions of Planck's constant: \[\mathrm{h} = \mathrm{M^1L^2T^{-1}}\]
04

Choose the correct option

Based on our calculation, the dimensional formula for Planck's constant is \(\mathrm{M^1L^2T^{-1}}\). Therefore, the correct option is: (b) \(\mathrm{M^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}}\)

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Most popular questions from this chapter

The radius of circle is \(1.26 \mathrm{~cm}\). According to the concept of significant figures area of it can be represented as - (a) \(4.9850 \mathrm{~cm}^{2}\) (b) \(4.985 \mathrm{~cm}^{2}\) (c) \(4.98 \mathrm{~cm}^{2}\) (d) \(9.98 \mathrm{~cm}^{2}\)

The resistance of two resistance wires are \(R_{1}=(100 \pm 5) \Omega\) and \(\mathrm{R}_{2}=(200 \pm 7) \Omega\) are connected in series. find the maximum absolute error in the equivalent resistance of the combination. (a) \(35 \Omega\) (b) \(12 \Omega\) (c) \(4 \Omega\) (d) \(9 \Omega\)

Dimensional formula for conductance is ........... (a) \(\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{2}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\) (d) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \mathrm{~A}^{2}\)

If \(\mathrm{A}=\mathrm{b}^{4}\) the fractional error in \(\mathrm{A}\) is $\ldots \ldots \ldots \ldots$ (a) \(\left[(\Delta \mathrm{b})^{4} /(\mathrm{b})\right]\) (b) \([\Delta \mathrm{b} / \mathrm{b}]\) (c) \(4[(\Delta \mathrm{b}) /(\mathrm{b})]\) (d) \((\Delta \mathrm{b})^{4}\)

The mass of substance is \(75.5 \mathrm{gm}\) and its volume is $25 \mathrm{~cm}^{3}$. It's density up to the correct significant figure is \(\ldots \ldots \ldots \ldots \ldots\) (a) \(3.02 \mathrm{gm} / \mathrm{cm}^{3}\) (b) \(3.200 \mathrm{gm} / \mathrm{cm}^{3}\) (c) \(3.02 \mathrm{gm} / \mathrm{cm}^{3}\) (d) \(3.1 \mathrm{gm} / \mathrm{cm}^{3}\)
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