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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

The density of a material in SI units is 128 kg m^{â€“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^{â€“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

2

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

3

The force of interaction between two atoms is given by F = $$\alpha $$$$\beta $$exp $$\left( { - {{{x^2}} \over {\alpha kt}}} \right)$$; where x is the distance, k is the Boltzmann constant and T is temperature and $$\alpha $$ and $$\beta $$ are two constants. The dimension of $$\beta $$ is :

A

M^{2}L^{2}T^{$$-$$2}

B

M^{2}LT^{$$-$$4}

C

MLT^{$$-$$4}

D

M^{0}L^{2}LT^{$$-$$4}

$$F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}$$

$$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$$

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

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

$$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$$

MLT^{$$-$$2} = M^{$$-$$1}T^{2}[$$\beta $$]

$$ \Rightarrow $$ [$$\beta $$] = M^{2}LT^{$$-$$4}

$$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$$

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

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

$$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$$

MLT

$$ \Rightarrow $$ [$$\beta $$] = M

4

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:

A

V^{$$-$$2}A^{2}F^{2}

B

V^{$$-$$4}A^{$$-$$2}F

C

V^{$$-$$4}A^{2}F

D

V^{$$-$$2}A^{2}F^{$$-$$2}

We know,

Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$

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

Let [Y] = [V]^{x} [A]^{y} [F]^{z}

$$ \therefore $$ [ ML^{-1}T^{-2}] =

[LT^{-1}]^{x} [LT^{-2}]^{y} [MLT^{-2}]^{z}

$$ \Rightarrow $$ [ ML^{-1}T^{-2}] =

[ M^{z} L^{x + y + z} T^{-x -2y - 2z}

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$$ \Rightarrow $$ x + y = -2 ........(1)

and -x -2y - 2z = -2

$$ \Rightarrow $$ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$$ \therefore $$ [Y] = V^{$$-$$4}A^{2}F^{1}

Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$

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

Let [Y] = [V]

$$ \therefore $$ [ ML

[LT

$$ \Rightarrow $$ [ ML

[ M

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$$ \Rightarrow $$ x + y = -2 ........(1)

and -x -2y - 2z = -2

$$ \Rightarrow $$ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$$ \therefore $$ [Y] = V

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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*

Impulse & Momentum *keyboard_arrow_right*

Rotational Motion *keyboard_arrow_right*

Gravitation *keyboard_arrow_right*

Properties of Matter *keyboard_arrow_right*

Heat and Thermodynamics *keyboard_arrow_right*

Waves *keyboard_arrow_right*

Vector Algebra *keyboard_arrow_right*

Electrostatics *keyboard_arrow_right*

Current Electricity *keyboard_arrow_right*

Magnetics *keyboard_arrow_right*

Alternating Current and Electromagnetic Induction *keyboard_arrow_right*

Ray & Wave Optics *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

Communication Systems *keyboard_arrow_right*

Practical Physics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*