IIT-JEE Advanced Physics:Topic 1.1 Unit of Physical Quantities: Important Questions

Question 

The density of a material in SI unit is 128 kg m-3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is:

(a) 40 (b) 16 (c) 640 (d) 410

Answer/Explanation

Ans

(a) 40
(a) Density of material in SI unit, = \(\frac{128kg}{m^3}\)

 

Question 

A metal sample carrying a current along X-axis with density J is subjected to a magnetic field Bz (along z-axis). The electric field EY developed along Y-axis is directly proportional to Jx as well as Bz. The constant of proportionality has SI unit

\((a)\;\frac{m^2}{A}\;(b)\;\frac{m^3}{As}\;(c)\;\frac{m^2}{As}\;(d)\;\frac{As}{m^3}\)

Answer/Explanation

Ans

(b)

(b) According to question

\(E_{y} \alpha J_{x}B_{Z}\)

\(\therefore \) Constant of proportionality

\(K=\frac{E_{y}}{B_{z}J_{x}}=\frac{C}{J_{x}}=\frac{m^3}{As}\)

 

Question

The SI unit of inductance, the henry can be written as

\((a)\;\frac{weber}{ampere}\;(b)\;\frac{volt-sec}{amp}\;(c)\;\frac{joule}{(amperer)^2}\;(d)\;ohm-second\)

Answer/Explanation

Ans:

(a, b, c, d)

Question

                                   Column I                                                                                     Column II

(A)     \(_{GM_eM_s}\) , \(_{G-}\)                                                           (p) (volt) (coulomb)(metre)
universal gravitational

constant,
\(_{M_e}\) – mass of the eart ,
\(_{M_s}\) – mass of the Su

(B)      \(_{\frac{3RT}{M}}\) , R-universal                                             (q) (kilogram) (metre)3(second)-2
gas constant,
T– absolute temperature,
M – molar mass

(C)     \(\frac{F^2}{q^2B^2}\) , F– Force ,                                              (r) (metre)2 (second)-2

             q-charge,

                       B-magnetic field

(D)     \(\frac{GM_e}{R_e}\) , G– universal ,                                         (s) (farad) (volt)2 (kg)-1

            Me – mass of the earth,
            Re – radius of the earth

Answer/Explanation

Answer

A → p, q

Using F= \(\frac{GM_eM_s}{r^2}\) ⇒ GMeMs = Fr2 = Nm2 = kg\(\frac{m}{s^2}\) x m2
=kg m3s-2
Also (volt) (coulomb) (metre) = (joule) (metre)

                                                                       (\(\because \) volt= Joule/coulomb)

                    = (Nm) (m) = Nm2 = kg m3s-2

B → r, s

Using \(\displaystyle v\)rms =  \(\sqrt{\frac{3RT}{M}}\) ⇒ \(v_{rms}^{2}\) = \(\frac{3RT}{M}\)

Unit of \(\frac{3RT}{M}\) is \(m^{2}\)  \(s^{-2}\)

Also (farad) (volt)2 (kg)-1 = (joule) kg-1  \((\because u = \frac{1}{2} cv^2)\)

          = N-m kg-1 = kg ms-2 m kg-1 = m2s-2

C → r, s

Using \(\displaystyle F = qvB \) ⇒ \(v^2= \frac{F^2}{q^2B^2}\)

∴  Unit of vis m2s-2 which is further equal to FV2 kg-1.

D → r, s

Reason : Escape velocity \(v_{e} = \sqrt{\frac{2GM}{R}}\) ⇒ \(v_{e}^{2} = \frac{2GM}{R}\)

∴  Unit of \(\frac{GM}{R}\) is m2 s-2

Question

                              Column I                                                                                     Column II

(A) Capacitance                                                                                           (i) ohm-second

(B) Inductance                                                                                            (ii) coulomb2-joule-1

(C) Magnetic Induction                                                                            (iii) coulomb (volt)-1

                                                                                                                       (iv) newton (amp-metre)-1

                                                                                                                       (v) volt-second(ampere)-1

Answer/Explanation

Answer              

Capacitance \(C=\frac{q}{v}=\frac{q}{\omega}\) coulomb-volt-1, coulomb2-joule-1.

Inductance \(\frac{L}{R}=t\)  and \(R=\frac{v}{I}\) ohm-sec, volt-second (ampere)-1

Magnetic Induction \(F=lIB\Rightarrow B=\frac{F}{lI}\) newton (ampere-metre)

Question

Give the MKS units for each of the following quantities.

(A) Young’s modulus

(B) Magnetic Induction

(C) Power of a lens

Answer/Explanation

Answer

(i) The M.K.S. unit of Young’s modulus \(\left ( Y=\frac{F}{A}/\frac{\triangle l}{l_0} \right )\)  is Nm-2.

(ii) The M.K.S. unit of magnetic induction \(\left ( B=\frac{\phi }{A} \right )\) is tesla or wb/m2.

(iii) The M.K.S. unit of power of lens \(P=\frac{1}{f\textup{(in metre)}}\) is dioptre or m-1

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