Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

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

$$\sqrt {{{Gh} \over {{c^3}}}} $$

Let t $$ \propto $$ G^{x} h^{y} c^{z}

$$ \therefore $$ [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 $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$

$$ \therefore $$ [t] = [G]

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

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 :

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

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

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

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

= 0.5 $$ \times $$ 10

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

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $$ \times $$ 0.5 $$ \times $$ 10

= 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

= 0.015 mm

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

= 5.725 mm

3

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

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,

n_{1} = 128

u_{1} = kg/m^{3}

And in new unit system,

n_{2} = ?

u_{2} = 50gm/(25cm)^{3}

As n_{1}u_{1} = n_{2}u_{2}

$$ \therefore $$ 128 $$ \times $$ (kg/m^{3}) = n_{2} $$ \times $$ 50gm/(25cm)^{3}

$$ \Rightarrow $$ 128 $$ \times $$ $${{1000\,gm} \over {{{\left( {100\,cm} \right)}^3}}}$$ = n_{2} $$ \times $$ $${{50\,gm} \over {{{\left( {25\,cm} \right)}^3}}}$$

$$ \Rightarrow $$ n_{2} = 128 $$ \times $$ $${{20} \over {{4^3}}}$$ = 40

density of a material in SI units is 128 kg m

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,

n

u

And in new unit system,

n

u

As n

$$ \therefore $$ 128 $$ \times $$ (kg/m

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

$$ \Rightarrow $$ n

4

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 cm^{3}

B

4264 $$ \pm $$ 81 cm^{3}

C

4300 $$ \pm $$ 80 cm^{3}

D

4260 $$ \pm $$ 80 cm^{3}

Volume of cylinder(V) = $$\pi $$r^{2}h

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

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

= 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

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

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

= 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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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

Work Power & Energy *keyboard_arrow_right*

Simple Harmonic Motion *keyboard_arrow_right*

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Vector Algebra *keyboard_arrow_right*

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Alternating Current and Electromagnetic Induction *keyboard_arrow_right*

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Electronic Devices *keyboard_arrow_right*

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Practical Physics *keyboard_arrow_right*