Q161JEE Main 2020MCQ

The dimension of stopping potential $V_0$ in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :[8 Jan 2020 Shift 1]

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📖 Explanation

$v_0 = h^xc ^yG^z A^w$ $\frac{ML^2T^{-2}}{AT}=(ML^2T^{-1})^x (LT^{-1})^y(M^{-1}L^{3}T^{-2})^zA^w$ $x = 0$ $z = -1$ $2\times 0 + y + 3x-1=2$ $y = 5$ $\Rightarrow v_0=h^0c^5G^{-1}A^{-1}$ So Bonus

Q162JEE Main 2020MCQ

If $I$ is moment of inertia, $F$ is force, $v$ is velocity, $E$ is energy and $L$ is length then, dimension of $\;\frac{ IFv ^{2}}{ EL ^{4}}$ will be:[4 Sep 2020 Shift 2]

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📖 Explanation

$\;\frac{ IFv ^{2}}{ EL ^{4}}$ $=\;\frac{( M ^{'} L ^{2})( M ^{'} L ^{'} T ^{-2})( L T ^{-1})^{2}}{( M ^{'} L ^{2} T ^{-2})( L ^{4})}$ $=\;\frac{ M ^{1} L ^{-2} T ^{-2}}{ L ^{3}}$ $= M ^{1} L ^{-1} T ^{-2}$

Q163JEE Main 2020MCQ

If momentum $(\ {P}),$ area $(\ {A})$ and time $(\ {T})$ aretaken to be the fundamental quantities then thedimensional formula for energy is :[2 Sep 2020 Shift 2]

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📖 Explanation

Let $[ {E}]=[ {P}]^{ {x}}[ {A}] {y}[ {T}]^{ {z}}$ ${ML}^{2} {T}^{-2}=\$ [{MLT}^{-1} ]^{ {x}} $[ {L}^{2} ]$ {y}[ {T}]^{ {z}}{ML}^{2} {T}^{-2}= {M}^{ {x}} {L}^{ {x}+2 {y}} {T}^{- {x}+ {z}}→{x}=1→{x}+2 {y}=21+2 y=2y= \frac{1}{2}→ -x+z=-2-1+z=-2z=-1[ {E}]=[{PA}^{1 / 2} {T}^{-1} ]$

Q164JEE Main 2020MCQ

Dimensional formula for thermal conductivity is (here $\ {K}$ denotes the temperature)[4 Sep 2020 Shift 1]

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📖 Explanation

$\because \frac{{d} \theta }{ {dt}}= {kA} \frac{{dT}}{ {dx}}$ ${k}={(\frac{{d} \theta }{ {dt}} )}/{ {A}(\frac{{dT}}{\ {dx}} )}$ $[ {k}]= \frac{[{ML}^{2} {T}^{-3} ]\$ }{[{L}^{2} ]$$[{KL}^{-1} ]$}=[{MLT}^{-3} {K}^{-1} ]$

Q165JEE Main 2020MCQ

The dimension of $B^2/{ 2\mu _0}$ , where B is magnetic field and $\mu _0$ is the magnetic permeability of vacuum, is:[7 Jan 2020 Shift 2]

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📖 Explanation

Magnetic energy stored per unit volume is $B^2/{2\mu _0}$ Dimension is $ML^{-1} T^{-2 }$

Q166JEE Main 2020MCQ

A physical quantity z depends on fourobservables $\ {a}, \ {b}, \ {c}$ and $\ {d},$ as $\ {z}={{a}^{2} \ {b}^{\frac{2}{3}}}/{\\sqrt{ {c}} {d}^{3}} .$ The percentage of error in the measurement of a, b, c and d $2 \%, 1.5 \%, 4 \%$ and $2.5 \%$ respectively. The percentage of error in z is:[5 Sep 2020 Shift 1]

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📖 Explanation

$\frac{\Delta Z}{Z}= \frac{2 \Delta a}{a}+ \frac{2}{3} \frac{\Delta b}{b}+ \frac{1}{2} \frac{\Deltac}{c}$ $+ \frac{3 \Deltad}{d}=14.5 \%$

Q167JEE Main 2019MCQ

The force of interaction between two atoms is given by F = αβ exp $(-x^2/{\alpha kt})$ ; where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is :[11 Jan 2019 Shift 1]

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📖 Explanation

F = αβ $e^(\frac{-x^2}{\alpha KT})$ $[x^2/{\alpha KT}]$ = $M^0L^0T^0$ $L^2/{[\alpha ]ML^2T^{-2}}$ = $M^0L^0T^0$ \Rightarrow [α] = $M^{-1}T^2$ [F] = [α] [β] $MLT^{-2}$ = $M^{-1}T^2$ [β] \Rightarrow [β] = $M^2LT^{-4}$

Q168JEE Main 2019MCQ

The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?[10 Jan 2019 Shift 2]

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📖 Explanation

${\Delta V}/V$ = 2 ${\Delta d}/d$ + ${\Delta h}/h$ = 2 $(\frac{0.1}{12.6})$ + $\frac{0.1}{34.2}$ V = 12.6 × $\pi /4$ × 314.2

Q169JEE Main 2019MCQ

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5µm diameter of wire is :[12 Jan 2019 Shift 1]

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📖 Explanation

Least count = $\frac{\text{Pitch}}{\text{Number} \text{of} \text{division} \text{on} \text{circular} \text{scale}}$ 5 × $10^{-6}$ = $10^{-3}/N$ N = 200

Q170JEE Main 2019MCQ

In SI units, the dimensions of $\sqrt{ \frac{\∊_{0}}{\mu_{0}}}$ is[8 Apr 2019 Shift 1]

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📖 Explanation

$\vec{P}_{1}\sqrt{ \frac{\∊_{0}}{\mu_{0}}}= \frac{1}{z}=$ Impedence $A^{2} T^{3} M^{-1} L^{-2}$

Q171JEE Main 2019MCQ

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :[9 Jan 2019 Shift 2]

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📖 Explanation

LC = $\frac{\text{Pitch}}{\text{No}. \text{of} \text{division}}$ LC = 0.5 × $10^{-2}$ mm +ve error = 3 × 0.5 × $10^{-2}$ mm = 1.5 × $10^{-2}$ mm = 0.015 mm Reading = MSR + CSR - (+ve error) = 5.5 mm + (48 × 0.5 × $10^{-2}$ ) - 0.015 = 5.5 + 0.24 - 0.015 = 5.725 mm

Q172JEE Main 2019MCQ

The density of a material in SI units 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 :[10 Jan 2019 Shift 1]

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📖 Explanation

${128kg}/m^3$ = $\frac{125(50g)(20)}{(25cm)^3(4)^3}$ = $\frac{128}{64}$ (20) units = 40 units

Q173JEE Main 2019MCQ

If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for Linear momentum would be:[8 Apr 2019 Shift 2]

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📖 Explanation

$P=S^{a} I^{b} h^{c}$ $M L T^{-1}= (M T^{-2} )^{a} (M L^{2} )^{b} (M L^{2} T^{-1} )^{c}$ $\Rightarrow P=S^{ \frac{1}{2}} I^{ \frac{1}{2}} h^{0}$

Q174JEE Main 2019MCQ

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :-[11 Jan 2019 Shift 2]

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📖 Explanation

$F/A$ = y . ${\Delta l}/l$ [Y] = $F/A$ Now from dimension F = ${ML}/T^2$ L = $F/M . T^2$ $L^2$ = $F^2/M^2 (V/A)^4$ Since T = $V/A$ $L^2$ = $F^2/{M^2A^2} v^4/A^2$ , F = MA $L^2$ = $V^4/A^2$ [Y] = $[F]/[A]$ = $F^1V^{-4}A^2$

Q175JEE Main 2019MCQ

Expression for time in terms of G (universalgravitational constant), h (Planck constant) andc (speed of light) is proportional to :[9 Jan 2019 Shift 2]

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📖 Explanation

F = ${GM^2}/R^2$ \Rightarrow G = $[M^{-1}L^3T^{-2}]$ E = hv \Rightarrow h = $[ML^2T^{-1}]$ C = $[LT^{-1}]$ t α $G^x h^y C^z$ [T] = $[M^{-1} L^3 T^{-2}]^x$ $[ML^2T^{-1}]^y$ $[LT^{-1}]^z$ $[M^0L^0T^1]$ = $[ $M^{-x}+ \,^{y}L^{3x+2y}$ + $^{z}T^{-2x} ^{-2x-y-z}$ ]$ on comparing the powers of M, L, T - x + y = 0 \Rightarrow x = y 3x + 2y + z = 0 ⇒ 5x + z = 0 ....(i) -2x - y - z = 1 ⇒ 3x + z = -1 ...(ii) on solving (i) & (ii) x = y = $1/2$ , z = - $5/2$ t α $\sqrt{{Gh}/c^5}$

Q176JEE Main 2019MCQ

A copper wire is stretched to make it 0.5%longer. The percentage change in its electricalresistance if its volume remains unchanged is : [9-Jan-2019 Shift 1]

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📖 Explanation

R = ${\rho l}/A$ and volume (V) = Al R = ${\rho l^2}/V$ \Rightarrow ${\Delta R}/R$ = ${2\Delta l}/l$ = 1 %

Q177JEE Main 2019MCQ

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:[8 Apr 2019 Shift 2]

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📖 Explanation

$\frac{d g}{g}=2 \frac{d T}{T}+ \frac{d l}{l}$ $\frac{d g}{g}=2 \times \frac{1}{30}+ \frac{10^{-1}}{55}$ $\%= \frac{200}{30}+ \frac{10}{55}=6.66+0.18=6.8 \%$

Q178JEE Main 2019MCQ

The area of a square is $5. 29cm ^{2}.$ The area of 7 such squares taking into account the significant figures is:[9 Apr 2019 Shift 2]

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📖 Explanation

Total area 5.29 ${cm}^{2} \\times 7$ $=37.03 {cm}^{2}$ In accord to significant figures it should be 37.0 cm $^{2}$

Q179JEE Main 2019MCQ

In the density measurement of a cube, the mass and edge length are measured as $(10.00 ±\ 0.10) \ kg$ and $( 0.10± 0.01 )\ m$ , respectively. The error in the measurement of density is:[9 Apr 2019 Shift 1]

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📖 Explanation

mass $=(10 ± 0.1) k g$ Length $=(0.1 \± 0.01) m$ Denisty $= \frac{m}{L^{3}}=D$ $\ =\ \frac{0.1}{10}+3 \times \frac{0.01}{0.1}$ $=0.01+0.3$ $=0.31 \ {kg} / {m}^{3}$

Q180JEE Main 2019MCQ

In the formula $X=5 Y Z^{2}, X$ and $Y$ , have dimensions of capacitance and magnetic field, respectively. what are the dimensions of Y in SI units?[10 Apr 2019 Shift 2]

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📖 Explanation

Given formula $X=5 Y Z^{2}$ $\Rightarrow[Y]= \$ [X Z^{-2}\ ]$=\ \frac{[X]}{[Z^{2}\ ]$}=\ \frac{\ $[M^{-1} L^{-2} T^{4} A^{2}\ ]$}{\ $[M^{2} L^{0} T^{-4} A^{-2}\ ]$}\Rightarrow[Y]= [M^{-3} L^{-2} T^{8} A^{4}\ ]$