Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If radius of the sphere is (5.3 $\pm$ 0.1)cm. Then percentage error in its volume will be

A. 3+6.01

\times \frac{100}{5.3}

\frac{1}{3} \times

0.01

\times \frac{100}{5.3}

(\frac{3 \times 0.1}{5.3}) \times 100

\frac{0.1}{5.3} \times 100

SOLUTION

Solution : C

Volume of sphere (v) = $\frac{4}{3} π γ^{3}$
Percentage error in volume =3 $\times \frac{Δ γ}{γ} \times$ 100 = (3 $\times \frac{0.1}{5.3}) \times$ 100

Question 2

The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, The maximum error in the measurement of pressure is

A. 1%

B. 2%

C. 6%

D. 8%

SOLUTION

Solution : D

$P = \frac{F}{A} = \frac{F}{l^{2}},$ So maximum error in pressure (P)
$(\frac{Δ P}{P} \times 100)_{m a x} = \frac{Δ F}{F} \times 100 + 2 \frac{Δ l}{1} \times 100$
$= 4 % + 2 \times 2 % = 8 %$

Question 3

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation $g = 4 π^{2} (\frac{l}{T^{2}})$ will be

10%

SOLUTION

Solution : C

Percentage error in $g = (% e r r o r i n l) + 2 (% e r r o r i n T) = 1 % + 2 (3 %) = 7 %$

Question 4

The length, breadth, and thickness of a block are given by l = 12 cm, b = 6 cm and t = 2.45 cm The volume of the block according to the idea of significant figures should be

A. 1

\times 10^{2} c m^{3}

B. 2

\times 10^{2} c m^{3}

C. 1.763

\times 10^{2} c m^{3}

D. None of these

SOLUTION

Solution : B

Volume V = l $\times$ b $\times$ t

= 12 $\times$ 6 $\times$ 2.45 = 176.4 $c m^{3}$

V = 176.4 $\times 10^{2} c m^{3}$

Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure. Hence, V = 2 $\times 10^{2} c m^{3} .$

Question 5

The position of a particle at time t is given by the relation x(t)= $(\frac{v_{0}}{α}) (1 - c^{- a t})$ where $v_{0}$ and $α$ are constants . The dimensions of $v_{0}$ and $α$ are respectively

M^{0} L T^{- 1}

and

T^{- 1}

M^{0} L^{1} T^{0}

and

T^{- 1}

M^{0} L^{1} T^{- 1}

and

L T^{- 2}

M^{0} L^{1} T^{- 1}

and T

SOLUTION

Solution : A

Dimension of $α$ t = $[M^{0} L^{0} T^{0}] ∴ [α] = [T^{- 1}]$
Again $[\frac{v_{0}}{α}]$ = [L] so $[v_{0}] = [L T^{- 1}]$

Question 6

The velocity v (in cm / sec) of a particle is given in terms of time t(in sec) by the relation v = $α$ t + $\frac{b}{t + c}$ ; the dimensions of a, b and c are

A. a =

L^{2}

, b = T, c =

L T^{2}

B. a =

L T^{2}

, b = LT, c = L

C. a =

L T^{- 2}

, b = L, c = T

D. a = L, b = LT, c =

T^{2}

SOLUTION

Solution : C

From the principle of dimensional homogeneity [v] = $[α t] \Rightarrow [α] = [L T^{- 2}]$ . similarly [b] = [L] and [c] = [T]

Question 7

From the equation $t a n θ = \frac{r g}{v^{2}}$ , one can obtain the angle of banking $θ$ for a cyclist taking a curve (the symbols have their usual meanings). Then say, it is

A. Both dimensionally and numerically correct

B. Neither numerically nor dimensionally correct

C. Dimensionally correct only

D. Numerically correct only

SOLUTION

Solution : C

Given equation is dimensionally correct because both sides are dimensionless but numerically wrong because the correct equation is $t a n θ = \frac{v^{2}}{r g} .$

Question 8

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity $η$ flowing per second through a tube of radius r and length l and having a pressure difference P across its end, is

A. V=

\frac{π P γ^{4}}{8 η l}

B. V=

\frac{π η l}{8 P γ^{4}}

C. V=

\frac{8 P η l}{π γ^{4}}

D. V =

\frac{π P η}{8 l γ^{4}}

SOLUTION

Solution : A

Formula for viscosity $η = \frac{π P γ^{4}}{8 V l} \Rightarrow$ V = $\frac{π P γ^{4}}{8 η l}$

Question 9

The SI unit of universal gas constant (R) is

A. Watt

K^{- 1} m o l^{- 1}

B. Newton

K^{- 1} m o l^{- 1}

C. Joule

K^{- 1} m o l^{- 1}

D. Erg

K^{- 1} m o l^{- 1}

SOLUTION

Solution : C

PV = nRT $\Rightarrow$ R = $\frac{P V}{n T}$ = $\frac{J o u l e}{m o l e \times K e l v i n} = J K^{- 1} m o l^{- 1}$

Question 10

From the dimensional consideration, which of the following equation is correct

A. T =

2 π \sqrt{\frac{R^{3}}{G M}}

B. T =2

π \sqrt{\frac{G M}{R^{3}}}

C. T =2

π \sqrt{\frac{G M}{R^{2}}}

D. T =2

π \sqrt{\frac{R^{2}}{G M}}

SOLUTION

Solution : A

By substituting the dimension in T = 2 $π \sqrt{\frac{R^{3}}{G M}}$
we get $\sqrt{\frac{L^{3}}{M^{- 1} L^{3} T^{- 2} \times M}}$ =T

Question 11

A force F is given by F=at + $b t^{2}$ , where t is time. What are the dimensions of a and b

M L T^{- 3}

and

M L^{2} T^{- 4}

M L T^{- 3}

and

M L T^{- 4}

M L T^{- 1}

and

M L T^{0}

M L T^{- 4}

and

M L T^{- 1}

SOLUTION

Solution : B

Dimension of a is same as that of $\frac{F}{t}$ .i.e. $M L T^{- 3}$

Dimension of b is same as that of $\frac{F}{t^{2}}$ .i.e $M L T^{- 4}$

Question 12

Dimensional formula $M L^{- 1} T^{- 2}$ does not represent the physical quantity

A. Young's modulus of elasticity

B. Stress

C. Strain

D. Pressure

SOLUTION

Solution : C

Strain $= \frac{Δ L}{L} \Rightarrow$ dimensionless quantity

Question 13

Select the pair whose dimensions are same

A. Pressure and stress

B. Stress and strain

C. Pressure and force

D. Power and force

SOLUTION

Solution : A

Pressure $= \frac{F o r c e}{A r e a} = M L^{- 1} T^{- 2}$
Stress $= \frac{R e s t o r i n g f o r c e}{A r e a} = M L^{- 1} T^{- 2}$

Question 14

The frequency of vibration f of a mass m suspended from a spring of spring constant K is given by a relation of this type f =C $m^{x} K^{y}$ ; where C is a dimensionless quantity. The value of x and y are

A. x =

\frac{1}{2}

, y =

\frac{1}{2}

B. x = -

\frac{1}{2}

, y = -

\frac{1}{2}

C. x =

\frac{1}{2}

, y = -

\frac{1}{2}

D. x = -

\frac{1}{2}

, y =

\frac{1}{2}

SOLUTION

Solution : D

By putting the dimensions of each quantity both sides we get $[T^{- 1}] = [M]^{x} [M T^{- 2}]^{y}$
Now comparing the dimenstions of quantities in both sides we get x+y =0 and 2y =1
$∴$ x = - $\frac{1}{2}$ , y = $\frac{1}{2}$

Question 15

A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $η$ . After some time the velocity of the ball attains a constant value known as terminal velocity $v_{r}$ . The terminal velocity depends on $(i)$ the mass of the ball $m, (i i) η, (i i i) r a n d (i v)$ acceleration due to gravity $g$ . Which of the following relations is dimensionally correct

v_{T} \propto \frac{m g}{η r}

v_{T} \propto \frac{η r}{m g}

v_{T} \propto η r m g

v_{T} \propto \frac{m g r}{η}

SOLUTION

Solution : A

By substituting dimension of each quantity in R.H.S. of option (a) we get $[\frac{m g}{η r}] = [\frac{M \times L T^{- 2}}{M L^{- 1} T^{- 1} \times L}]$
$= [L T^{- 1}] .$
This option gives the dimension of velocity.

Question 16

If velocity v, acceleration A and force B are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be

F A^{- 1} v

F v^{3} A^{- 2}

F v^{2} A^{- 1}

F^{2} v^{2} A^{- 1}

SOLUTION

Solution : B

L $\propto v^{x} A^{y} F^{z} \Rightarrow$ L = $k v^{x} A^{y} F^{z}$
Putting the dimenstions in the above relation
$[M L^{2} T^{- 1}] = k [L T^{- 1}]^{x} [L T^{- 2}]^{y} [M L T^{- 2}]^{z}$

$\Rightarrow [M L^{2} T^{- 1}] = k [M^{z} L^{x + y + z} T^{- x - 2 y - 2 z}]$
Comparing the pwers of M,L andT
z=1 ...(i)
x+y+z =2...(ii)
-x-2y-2z =-1 ...(iii)
On solving (i),(ii), and (iii) x=3, y=-2 ,z=1
So dimension of L in terms of v,A and f
[L] = $[F v^{3} A^{- 2}]$

Question 17

Dimensions of coefficient of viscosity are

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

M L^{2} T^{- 1}

M L^{- 1} T^{- 1}

M L T

SOLUTION

Solution : C

$F = - η . A \frac{d v}{d x} \Rightarrow [η] = [M L^{- 1} T^{- 1}]$ where A is the area, v the velocity and dx the change in displacement.

Question 18

Two quantities A and B have different dimensions. Which mathematical operation given below is physically meaningful

\frac{A}{B}

A + B

A - B

D. can be A and B

SOLUTION

Solution : A

Quantities having different dimensions can only be divided or multiplied but they cannot be added or subtracted.

Question 19

In C.G.S. system the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is

A. 0.036

B. 0.36

C. 3.6

D. 36

SOLUTION

Solution : C

$n_{2} = n_{1} {(\frac{M_{1}}{M_{2}})}^{1} {(\frac{L_{1}}{L_{2}})}^{1} {(\frac{T}{T_{2}})}^{- 2}$
$= 100 {(\frac{g m}{k g})}^{1} {(\frac{c m}{m})}^{1} {(\frac{s e c}{m i n})}^{- 2}$
$= 100 {(\frac{g m}{10^{3} g m})}^{1} {(\frac{c m}{10^{2} c m})}^{1} {(\frac{s e c}{60 s e c})}^{- 2}$
$n_{2} = \frac{3600}{10^{3}} = 3.6$

Question 20

The unit of Stefan's constant $σ$ is

W m^{- 2} K^{- 1}

W m^{2} K^{- 4}

W m^{- 2} K^{- 4}

W m^{- 2} K^{4}

SOLUTION

Solution : C

Stefan's law is $E = σ (T^{4}) \Rightarrow σ = \frac{E}{T^{4}}$
where, $E = \frac{E n e r g y}{A r e a \times T i m e} = \frac{W a t t}{m^{2}}$
$σ = \frac{W a t t - m^{- 2}}{K^{4}} = W a t t = m^{- 2} K^{- 4}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Question 16

SOLUTION

Question 17

SOLUTION

Question 18

SOLUTION

Question 19

SOLUTION

Question 20

SOLUTION

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