Sheet 03 Relation between different Physical Quantities.
The force\(\left(F\right)\) acting on an object depends upon, (i) Mass of the object \(\left(m\right)\) and (ii)Acceleration of the object \(\left(a\right)\) . Find: the expression of force \(\left(F\right)\) by the method of dimension.
Ans: \(F=kma\).
The kinetic energy possessed by an object depends upon (i) Mass \(\left(m\right)\) (ii) Speed \(\left(v\right)\). Find; the expression for kinetic energy of an object in motion using method of dimensions.
Ans: \(K.E=Kmv^{2}\).
By employing the method of dimensions derive the formula \(S=ut+\frac{1}{2}at^{2}\). Where; u = Initial Velocity, S = Distance Travelled in time t and a is acceleration of the object.
Ans: \(S=K_{1}u.t+K_{2}a.t^{2}\).
A planet moves around its sun in a circular orbit. Assume that the period of revolution is \(T\) of the planet depends upon radius \(\left(R\right)\) of its orbit, mass of the sun \(\left(M\right)\) and universal gravitational constant \(\left(G\right)\) then prove dimensionally; \(T^{2}\propto\frac{R^{3}}{GM}\).
Ans: \(t=2\pi\sqrt{\frac{R^{3}}{GM}}\).
The viscosity of a gas \(\left(\eta\right)\) depends on mass \(\left(m\right)\), the effective diameter \(\left(d\right)\) and the mean speed of molecules \(\left(v\right)\). Use the dimensional analysis to find \(\eta\) as a function of these variables.
Ans: \(\eta=K\frac{mv}{d^{2}}\).
Assume that the volume of the liquid flowing per second or (Rate of flow of liquid) through a narrow pipe depends upon, (i) The coefficient of viscosity \(\left(\eta\right)\) of the liquid. (ii) Radius \(\left(r\right)\) of the pipe. (iii) Pressure gradient \(\left(\frac{p}{l}\right)\) along the pipe. Show that: by the method of dimensions that how rate of flow of liquid depends on these quantities?
Ans: \(V=\frac{\pi}{8}\frac{pr^{4}}{\eta l}\).
A body of mass \(m\) hung at one end of the spring executes S.H.M. The force constant of spring is \(K\) while its time period of vibrations is \(T\). Prove by the method of dimensions that \(T=\frac{2\pi m}{K}\) is incorrect. Derive the correct relation for time period by the using the method of dimensions.
Ans: \(T=2\pi\sqrt{\frac{m}{K}}\).
Suppose; that the oscillation of a simple pendulum depends on, (i) Mass of the bob \(\left(m\right)\). (ii) Effective length \(\left(l\right)\) of the pendulum. (iii) Acceleration due to gravity \(\left(g\right)\). and (iv) Angular displacement \(\left(\theta\right)\). Derive an equation for time period of the simple pendulum using dimensional analysis.
Ans: \(T=K\sqrt{\frac{l}{g}}\).
A small steel ball of radius \(\left(r\right)\) falling under gravity through a viscous liquid of coefficient of viscosity \(\left(\eta\right)\) attains a constant velocity \(\left(v\right)\). The velocity depends upon the weight \(\left(Mg\right)\) of the ball, coefficient of viscosity \(\left(\eta\right)\) and the radius \(\left(r\right)\) of the ball. Using dimensional methode find the expression for \(v\).
Ans: \(v=\frac{KMg}{\eta r}\).
Given: that the time period of oscillation of a water bubble from an explosion under water depends on; (i) Static pressure \(\left(p\right)\) . (ii) Density of the water \(\left(\rho\right)\) . and (iii) Total energy of the explosion \(\left(E\right)\). Using the method of dimensions derive an expression for time period \(\left(T\right)\) of the oscillation of the water bubble.
The frequency of vibrations \(\left(\nu\right)\) of a string having mass \(\left(M\right)\) depends on length of string \(\left(l\right)\), tension in string \(\left(T\right)\) and mass per unit length \(\left(m\right)\) of string. Use the methode of dimension to establish the formula for frequency. Value of dimensionless constnat \(=\frac{1}{2}\).
Ans: \(\nu=\frac{1}{2l}\sqrt{\frac{T}{m}}\).
Obtain: an expression for centrifugal force \(\left(F\right)\) acting on a particle of mass \(\left(m\right)\) moving with a velocity \(\left(v\right)\) in a circle of radius \(\left(r\right)\) then prove dimensionally; \(F\propto\frac{mv^{2}}{r}\).
