Units and Measurements Test

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Units and Measurements Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

The number of significant figure in the result of (5.0m+6.0m) is

A
Two

B
Three

C
Four

D
One

Solution

$$\begin{array}{l}\text { Significant figure in } 5.0+6.0=13.0 \\\text { So, the answer is three because zero after } \\\text { decimal is also counted. }\end{array}$$
Question 2
1 / -0

$$\left[M^{1}L^{2}T^{-3}\right]$$ are the dimensions of ....

A
pressure

B
power

C
force

D
work

Solution

$$Power=\dfrac{Work}{Time}=\dfrac{M^1 L^2T^{-2}}{T}=M^1L^2T^{-3}$$
Question 3
1 / -0

If % error in length, diameter, current and voltage are same than which of the following affects % error in measurement of resistivity, the most:

A
length measurement

B
voltage measurement

C
current measurement

D
diameter measurement

Solution

$$R = \frac{{\rho l}}{\Delta } \Rightarrow \rho = \frac{{R\Delta }}{l}$$
$$\rho = \frac{{R\left( {M{r^2}} \right)}}{l}$$
$$\therefore $$ Diameter measurement is most affect the error of $$\rho $$ $$($$Resistivity$$)$$ because it has power of two$$.$$
Hence,
option $$(D)$$ is correct answer.
Question 4
1 / -0

The dimension of $$ \cfrac {1}{2}\epsilon_o E^2$$ permittivity of free space and E:
Intensity of electric field) is

A
$$[MLT^{-1}]$$

B
$$[ML^2T^{-1}]$$

C
$$[ML^{-1}T^{-2}]$$

D
$$[ML^2T^{-2}]$$

Solution

Energy density of an electric field $$E$$ is ,

$$\mu_E=\dfrac 12 \epsilon_0 E^2$$

where $$\epsilon_0$$ is permitivity of free space

$$\mu_E=\dfrac{Energy}{Volume}=\dfrac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}]$$

Hence, the dimension of $$\dfrac 12 \epsilon_0 E^2$$ is $$[ML^{-1}T^{-2}]$$
Question 5
1 / -0

The radius of a sphere is measured as $$(10 \pm 0.02\ \% )$$ cm. The error in the measurement of its volume is

A
$$25.1$$ cc

B
$$25.12$$ cc

C
$$2.51$$ cc

D
$$251.2$$ cc

Solution

Given,
$$r=10\pm 0.02$$%
Volume, $$V=\dfrac{4}{3}\pi r^3$$. . . . . . . . .(1)
$$\dfrac{\Delta V}{V}=3\dfrac{\Delta r}{r}$$
$$\Delta V=3V\dfrac{\Delta r}{r}$$
$$\Delta V=\dfrac{4}{3}\pi r^3\times 3\times \dfrac{\Delta r}{r}$$
$$\Delta V=\dfrac{4}{3}\times 3.14\times 10\times 10\times 10\times 3\times \dfrac{0.02}{10}=25.12cc$$
The correct option is B.
Question 6
1 / -0

The dimensional formula for plank's constant and angular momentum are

A
$$\left[ M{ L }^{ 2 }{ T }^{ -2 } \right] and \left[ ML{ T }^{ -1 } \right] $$

B
$$\left[ M{ L }^{ 2 }{ T }^{ -1 } \right] and \left[ ML^{ 2 }{ T }^{ -1 } \right] $$

C
$$\left[ M{ L }^{ 3 }{ T }^{ 1 } \right] and \left[ ML^{ 2 }{ T }^{ -2 } \right] $$

D
$$\left[ M{ L }^{ 2 }{ T }^{ -1 } \right] and \left[ ML^{ 2 }{ T }^{ -2 } \right] $$

Solution

Planck's constant, symbolized h, relates the energy in one quantum (photon) of electromagnetic radiation to the frequency of that radiation.

$$h=\dfrac Ev=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$$

Angular momentum $$=$$ Moment of inertia $$\times$$ Angular velocity (Angular momentum)

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

So, correct answer is option (B).
Question 7
1 / -0

The velocity of a particle varies with distance x from a fixed origin as $$v=Ax+\dfrac { { Bx }^{ 2 } }{ C+x } $$, where A,B and C are dimensional constant then the dimensional formula of $$\dfrac { AB }{ C } $$ is

A
$$\left[ ML{ T }^{ -2 } \right] $$

B
$$\left[ M^{ 0 }L^{ -1 }{ T }^{ -2 } \right] $$

C
$$\left[ ML^{ -1 }{ T }^{ -2 } \right] $$

D
$$\left[ M^{ 0 }L^{ 0 }{ T }^{ -2 } \right] $$
Question 8
1 / -0

Given $$m$$ is the mass per unit length, $$T$$ is the tension $$\& L$$ is the length of the wire the dimensional formula for $$L \left( \cfrac { m } { T } \right) ^ { 1 / 2 }$$ is same as that for:-

A
Wavelength

B
velocity

C
Time period

D
Frequency

Solution
Question 9
1 / -0

The physical quantity having the dimensions$$\left[ { M }^{ -1 }{ L }^{ -3 }{ T }^{ 3 }{ A }^{ 2 } \right] $$ IS

A
resistance

B
resistivity

C
electrical conductivity

D
electromotive force

Solution
Question 10
1 / -0

Dimensions of electrical resistance are :

A
$$[ML^2T^{-3}A^{-1}]$$

B
$$[ML^2T^{-3}A^{-2}]$$

C
$$[ML^3T^{-3}A^{-2}]$$

D
$$[ML^2T^{3}A^{2}]$$

Solution

Since according to Ohm's Law :

$$V=IR$$

Hence, $$R=\dfrac {V}{I}$$

Where $$V=\dfrac {W}{q}=\dfrac {work\ done}{charge}$$

Hence; Potential $$=\dfrac {[M^1 L^2 T^{-2}]}{[AT]}$$

$$\therefore \ [V]=[M^1 L^2 T^{-3} A^{-1}]$$

From above equations,

$$\therefore \ $$ Resistance $$=[R]=\dfrac {[M^1 L^2 T^{-3} A^{-1}]}{[A]}$$

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

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Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 74

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Units and Measurements Test - 74

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SHARING IS CARING

Question 1

Two

Three

Four

One

Solution

Question 2

pressure

power

force

work

Solution

Question 3

length measurement

voltage measurement

current measurement

diameter measurement

Solution

Question 4

$$[MLT^{-1}]$$

$$[ML^2T^{-1}]$$

$$[ML^{-1}T^{-2}]$$

$$[ML^2T^{-2}]$$

Solution

Question 5

$$25.1$$ cc

$$25.12$$ cc

$$2.51$$ cc

$$251.2$$ cc

Solution

Question 6

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

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

$$\left[ M{ L }^{ 3 }{ T }^{ 1 } \right] and \left[ ML^{ 2 }{ T }^{ -2 } \right] $$

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

Solution

Question 7

$$\left[ ML{ T }^{ -2 } \right] $$

$$\left[ M^{ 0 }L^{ -1 }{ T }^{ -2 } \right] $$

$$\left[ ML^{ -1 }{ T }^{ -2 } \right] $$

$$\left[ M^{ 0 }L^{ 0 }{ T }^{ -2 } \right] $$

Question 8

Wavelength

velocity

Time period

Frequency

Solution

Question 9

resistance

resistivity

electrical conductivity

electromotive force

Solution

Question 10

$$[ML^2T^{-3}A^{-1}]$$

$$[ML^2T^{-3}A^{-2}]$$

$$[ML^3T^{-3}A^{-2}]$$

$$[ML^2T^{3}A^{2}]$$

Solution

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