151. Equation of force \(F=at+bt^{2}\) where \(F\) is force in Newton, \(t\) is time in second, then write unit of \(b\).
A. \(Nm^{-1}\)
B. \(Nm^{2}\)
C. \(Nm\)
D. \(Nm^{-2}\)
Answer : Option D
Explanation :
\(F=at+bt^{2}\)
\(F=bt^{2}=at\)
\(b=\frac{F}{t^{2}}=\frac{N}{m^{2}}\)
152. Pressure \(P=\frac{at^{2}}{bx}\) where x = distance, t= time find the dimensional formula for \(\frac{a}{b}\)
A. \(M^{1}L^{0}T^{-4}\)
B. \(M^{1}L^{1}T^{-1}\)
C. \(M^{1}L^{0}T^{-2}\)
D. \(M^{1}L^{0}T^{2}\)
Answer : Option A
153. \(F=A_{0}\left ( 1-e^{-Bxt^{2}} \right )\) where F is force and x is displacement. write the dimension formula of B
A. \(M^{2}L^{1}T^{-1}\)
B. \(M^{0}L^{-1}T^{-2}\)
C. \(M^{1}L^{0}T^{-2}\)
D. \(M^{1}L^{2}T^{-1}\)
Answer : Option B
Explanation :
\(F=A_{0}\left ( 1-e^{-Bxt^{2}} \right )\)
\(Bxt^{2}=dimentionless\)
\(B=\frac{M^{0}L^{0}T^{0}}{xt^{2}}=M^{0}L^{-1}T^{-2}\)
154. Equation of physical quantity \(v=at+bt^{2}\) where v = velocity, t = time so write the dimensional formula of a in this equation
A. \(M^{0}L^{1}T^{-1}\)
B. \(M^{1}L^{1}T^{-1}\)
C. \(M^{0}L^{1}T^{-2}\)
D. \(M^{1}L^{2}T^{0}\)
Answer : Option C
155. The resistivity of resistive wire is \(\rho =\frac{AR}{L}\) where L = length of wire, A = Area of wire and R is resistance of wire find dimension formula of \(\rho \)
A. \(M^{1}L^{3}T^{-3}A^{-2}\)
B. \(M^{1}L^{2}T^{-3}A^{-2}\)
C. \(M^{2}L^{3}T^{1}A^{2}\)
D. \(M^{2}L^{3}T^{-3}A^{-2}\)
Answer : Option A
156. Density of substance in CGS system is \(3.125\ gm/cm^{3}\) what is its magnitude in SI system ?
A. 0.3125
B. 3.125
C. 31.25
D. 3125
Answer : Option D
Explanation :
\(Density=3.125\times \frac{gm}{cm^{3}}\)
\(=\frac{3.125\times 10^{-3}kg}{10^{-6}m^{3}}\)
\(3125\ kg/m^{3}\)
157. A cube has numerically equal volume and surface area calculate the volume of such a cube.
A. 2000 Unit
B. 216 Unit
C. 2160 Unit
D. 1000 Unit
Answer : Option B
Explanation :
Volume of cube \(V=a^{3}\)
total surface area of cube \(A=6a^{2}\)
\(\therefore V=A\)
\(a^{3}=6a^{2}\)
\(a=6\)
\(\therefore V=6^{3}=216\ unit\)
158. Which out of the following is dimensionally correct.
A. \(p^{2}=h\rho g\)
B. \(p=h\rho ^{2}g\)
C. \(p=h\rho g\)
D. \(p=h^{2}\rho g\)
Answer : Option C
159. If energy \(E=G^{p}h^{q}c^{r}\) where G is the universal gravitational constant. h is the plank’s constant and c is the velocity of light, then the values of p, q and r are respectively
A. \(-\frac{1}{2},\frac{1}{2},\frac{5}{2}\)
B. \(\frac{1}{2},\frac{1}{2},\frac{5}{2}\)
C. \(\frac{5}{2},\frac{1}{2},-\frac{1}{2}\)
D. \(\frac{1}{2},-\frac{1}{2},\frac{5}{2}\)
Answer : Option A
Explanation :
\(E=G^{p}h^{q}c^{r}\)
\(E=M^{1}L^{2}T^{-2}\)
\(G=M^{-1}L^{3}T^{-2}\)
\(h=M^{1}L^{2}T^{-1}\)
\(c=M^{0}L^{1}T^{-1}\) take it
\(\left ( M^{1}L^{2}T^{-2} \right )=\left ( M^{-1}L^{3}T^{-2} \right )^{p}\left ( M^{1}L^{2}T^{-1} \right )^{q}\left ( M^{0}L^{1}T^{-1} \right )^{r}\)
\(=M^{-p+q}L^{3p+2q+r}T^{-2p-q-r}\)
\(\therefore p=-\frac{1}{2},q=\frac{1}{2},r=\frac{5}{2}\)
160. If the centripetal force is of the form \(m^{a}v^{b}r^{c}\) find the values of a, b and c
A. 1,2,1
B. 1,2,–1
C. 1,3,–2
D. –1,3,–1
Answer : Option B
Explanation :
\(F\ \alpha\ m^{a}v^{b}r^{c}\)
\(F=M^{1}L^{1}T^{-2}\)
\(v=M^{0}L^{1}T^{-1}\)
\(r=M^{0}L^{1}T^{0}\)
\(m=M^{1}L^{0}T^{0}\) take it
\(\left ( M^{1}L^{1}T^{-2} \right )=\left ( M^{1} \right )^{a}\left ( L^{1}T^{-1} \right )^{b}\left ( L^{1} \right )^{c}\)
\(=M^{a}L^{b+c}T^{-b}\)
\(\therefore a=1,b=2,c=-1\)