141.
Assuming that the mass $$m$$ of the largest stone that can be moved by a flowing river depends upon the velocity $$v$$ of the water, its density $$\rho $$ and the acceleration due to gravity $$g.$$ Then m is directly proportional to :
142.
According to Newton, the viscous force acting between liquid layers of area $$A$$ and velocity gradient $$\frac{{\Delta v}}{{\Delta z}}$$ is given by $$F = - \eta A\frac{{dv}}{{dz}},$$ where $$\eta $$ is constant called
143.
The density of a sphere is measured by measuring its mass and diameter. If, it is known that the maximum percentage errors in the measurement are $$2\% $$ and $$3\% ,$$ then find the maximum percentage error in the measurement of density?
Let $$m$$ and $$d$$ be the mass and diameter of the sphere, then the density $$\rho $$ of the sphere is given by
$$\eqalign{
& \rho = \frac{{{\text{mass}}}}{{{\text{volume}}}} = \frac{m}{{\frac{4}{3}\pi {{\left( {\frac{d}{2}} \right)}^3}}} = \frac{{6m}}{{\pi {d^3}}} \cr
& {\text{or}}\,\,{\left( {\frac{{d\rho }}{\rho } \times 100} \right)_{\max }} = \left| {\frac{{dm}}{m} \times 100} \right| + \left| {\frac{{3d\left( d \right)}}{d} \times 100} \right| \cr
& = 2 + 3\left( 3 \right) \cr
& = 11\% \cr} $$
144.
The dimension of $$\left( {\frac{1}{2}} \right){\varepsilon _0}{E^2}$$ ($${\varepsilon _0}$$ : permittivity of free space, $$E$$ electric field)
Note: Here $$\left( {\frac{1}{2}} \right){e_0}{E^2}$$ represents energy per unit volume.
$$\left[ {{e_0}} \right]\left[ {{E^2}} \right] = \frac{{\left[ {{\text{Energy}}} \right]}}{{\left[ {{\text{volume}}} \right]}} = \frac{{M{L^2}{T^{ - 2}}}}{{{L^3}}} = M{L^{ - 1}}{T^{ - 2}}$$
145.
In an experiment four quantities $$a,b,c$$ and $$d$$ are measured with percentage error $$1\% ,2\% ,3\% $$ and $$4\% $$ respectively. Quantity $$P$$ is calculated as follows
$$P = \frac{{{a^3}{b^2}}}{{cd}}\% $$ error in $$P$$ is
146.
If force $$\left( F \right),$$ velocity $$\left( V \right)$$ and time $$\left( T \right)$$ are taken as fundamental units, then the dimensions of mass are
$$\eqalign{
& F = \frac{{GMm}}{{{R^2}}} \cr
& \therefore G = \frac{{F{R^2}}}{{Mm}} \Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] \cr} $$
150.
In the determination of Young’s modulus $$\left( {Y = \frac{{4MLg}}{{\pi l{d^2}}}} \right)$$ by using Searle’s method, a wire of length $$L = 2 \,m$$ and diameter $$d=0.5 \,mm$$ is used. For a load $$M= 2.5 \,kg,$$ an extension $$l = 0.25\,mm$$ in the length of the wire is observed. Quantities $$d$$ and $$l$$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $$0.5 \,mm.$$ The number of divisions on their circular scale is $$100.$$ The contributions to the maximum probable error of the $$Y$$ measurement-
A
due to the errors in the measurements of $$d$$ and $$l$$ are the same.
B
due to the error in the measurement of d is twice that due to the error in the measurement of $$l.$$
C
due to the error in the measurement of $$l$$ is twice that due to the error in the measurement of $$d.$$
D
due to the error in the measurement of $$d$$ is four times that due to the error in the measurement of $$l.$$
Answer :
due to the errors in the measurements of $$d$$ and $$l$$ are the same.