The time period of a body undergoing simple h | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.The dimensional formula for pressure $p$ is $[M^1 L^{-1} T^{-2}]$.The dimensional f

Vedclass · Accepted Answer

$-\frac{3}{2}, \frac{1}{2}, 1$

Vedclass · Answer

(D) The dimensional formula for time period $T$ is $[M^0 L^0 T^1]$.The dimensional formula for pressure $p$ is $[M^1 L^{-1} T^{-2}]$.The dimensional formula for density $D$ is $[M^1 L^{-3} T^0]$.The dimensional formula for surface tension $S$ is $[M^1 L^0 T^{-2}]$.Given the relation $T = p^a D^b S^c$,we equate the dimensions on both sides:$[M^0 L^0 T^1] = [M^1 L^{-1} T^{-2}]^a [M^1 L^{-3} T^0]^b [M^1 L^0 T^{-2}]^c$$[M^0 L^0 T^1] = [M^{a+b+c} L^{-a-3b} T^{-2a-2c}]$Comparing the powers of $M, L,$ and $T$ on both sides:$1$) $a + b + c = 0$$2$) $-a - 3b = 0 \implies a = -3b$$3$) $-2a - 2c = 1$Substitute $a = -3b$ into equation $(1)$:$-3b + b + c = 0 \implies c = 2b$Substitute $a = -3b$ and $c = 2b$ into equation $(3)$:$-2(-3b) - 2(2b) = 1$$6b - 4b = 1 \implies 2b = 1 \implies b = \frac{1}{2}$Now,find $a$ and $c$:$a = -3b = -3(\frac{1}{2}) = -\frac{3}{2}$$c = 2b = 2(\frac{1}{2}) = 1$Thus,the values are $a = -\frac{3}{2}, b = \frac{1}{2}, c = 1$.

The time period of a body undergoing simple harmonic motion is given by $T=p^{a} D^{b} S^{c}$,where $p$ is the pressure,$D$ is density,and $S$ is surface tension. The values of $a, b,$ and $c$ respectively are

Similar Questions

The potential energy of a point particle is given by the expression $V(x) = -\alpha x + \beta \sin(x / \gamma)$. $A$ dimensionless combination of the constants $\alpha, \beta$ and $\gamma$ is

If speed of light in vacuum $\left(3 \times 10^8 \,m/s\right)$, acceleration due to gravity $\left(10 \,m/s^2\right)$ and mass of electron $\left(9.1 \times 10^{-31} \,kg\right)$ are taken as fundamental physical quantities, then the unit of time in this system is

If the orbital velocity of a planet is given by $v = G^a M^b R^c$,then:

In the relation $y = a \cos (\omega t - kx)$,the dimensional formula for $k$ is

In a system,the unit of mass is $A \,kg$,length is $B \,m$,and time is $C \,s$. Then,the value of $10 \,N$ in this system is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module