1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
A
$$\sqrt {{{h{c^5}} \over G}} $$
B
$$\sqrt {{{{c^3}} \over {Gh}}} $$
C
$$\sqrt {{{Gh} \over {{c^5}}}} $$
D
$$\sqrt {{{Gh} \over {{c^3}}}} $$

Explanation

Let t $$ \propto $$ Gx hy cz

$$ \therefore $$   [t] = [G]x [h]y [c]z    . . . . . (1)

We know,

F = $${{G{M^2}} \over {{R^2}}}$$

$$ \Rightarrow $$  G = $${{F{R^2}} \over {{M^2}}}$$

$$ \therefore $$  [G] = $${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$$

[G] = $$[{M^{ - 1}}{L^3}{T^{ - 2}}]$$

Also,

E = hf

$$ \therefore $$  [h] = $${{[E]} \over {[F]}}$$

= $${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$$

= $$\left[ {M{L^2}{T^{ - 1}}} \right]$$

[C] = $$\left[ {{M^o}L\,{T^{ - 1}}} \right]$$

From equation (1) we get,

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$$

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$$

By comparing the power of M, L, T

$$-$$ x + y = 0

$$ \Rightarrow $$  x = y

3x + 2y + z = 0

$$ \Rightarrow $$  5x + z = 0 . . . . . (2)

$$-$$ 2x $$-$$ y $$-$$ z = 1

$$ \Rightarrow $$  $$-$$ 3x $$-$$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = $${1 \over 2}$$ = y and z = $$-$$ $${5 \over 2}$$

$$ \therefore $$  t $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

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 :
A
5.755 mm
B
5.950 mm
C
5.725 mm
D
5.740 mm

Explanation

We know,

Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$

$$ \therefore $$  LC = $${{0.5} \over {100}}$$

= 0.5 $$ \times $$ 10$$-$$2 mm

Reading = MSR + CSR $$-$$ positive error

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$$ \therefore $$  positive error

= 3 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm

= 0.015 mm

$$ \therefore $$  Reading = 5.5 + 0.24 $$-$$ 0.015

= 5.725 mm
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

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 -
A
40
B
640
C
16
D
410

Explanation

Here given that

density of a material in SI units is 128 kg m–3

And a new unit system is introduced where 1 unit of length = 25 cm and 1 unit of mass = 50 g

You should know that, physical quantity is same in any unit system. And to calculate a physical quantity yo should know two things

(1) numerical value of the physical quantity (n)

(2) unit of the physical quantity (u)

And n $$ \times $$ u = constant in any unit system.

Here in SI unit system,

n1 = 128

u1 = kg/m3

And in new unit system,

n2 = ?

u2 = 50gm/(25cm)3

As n1u1 = n2u2

$$ \therefore $$ 128 $$ \times $$ (kg/m3) = n2 $$ \times $$ 50gm/(25cm)3

$$ \Rightarrow $$ 128 $$ \times $$ $${{1000\,gm} \over {{{\left( {100\,cm} \right)}^3}}}$$ = n2 $$ \times $$ $${{50\,gm} \over {{{\left( {25\,cm} \right)}^3}}}$$

$$ \Rightarrow $$ n2 = 128 $$ \times $$ $${{20} \over {{4^3}}}$$ = 40
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

The diameter and height of a cylinder are measured by a meter scale to be 12.6 $$ \pm $$ 0.1 cm and 34.2 $$ \pm $$ 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?
A
4264.4 $$ \pm $$ 81.0 cm3
B
4264 $$ \pm $$ 81 cm3
C
4300 $$ \pm $$ 80 cm3
D
4260 $$ \pm $$ 80 cm3

Explanation

Volume of cylinder(V) = $$\pi $$r2h

= $$\pi {{{d^2}} \over 4}h$$

= $$3.14 \times {{{{\left( {12.6} \right)}^2}} \over 4} \times 34.2$$

= 4260

$${{\Delta V} \over V} = 2{{\Delta d} \over d} + {{\Delta h} \over h} = 2\left( {{{0.1} \over {12.6}}} \right) + {{0.1} \over {34.2}}$$ = 0.0188

$$ \therefore $$ $$\Delta $$V = 0.0188 $$ \times $$ 4260 = 80

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