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CBSE Questions for Class 11 Engineering Physics Units And Measurement Quiz 16 - MCQExams.com
CBSE
Class 11 Engineering Physics
Units And Measurement
Quiz 16
Dimensional formula of the product of the physical quantity P and Q is $$\left[ { M }^{ 1}{ L }^{ 2 }{ T }^{ -2 } \right] $$ and the dimensional formula of $$\dfrac { P }{ Q } $$ is $$\left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 } \right] $$ then
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'P' represents work
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'Q' represent velocity
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'P' represents force
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'Q' represents displacement
If the speed of light (c) , acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities , then the dimensions of length will be
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$$ c^2g^{-1}p^0 $$
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$$ cg^0p^{-1} $$
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$$ c^{-1}gp^0 $$
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$$ cg^{-1}p^0 $$
Characteristics of Scientific observation are.
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Accuracy
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Precision and objectivity
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Systematic and recorded
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All of the above
The distance covered by a particle in time $$t$$ is given by$$ x = a + bt + ct^2 + dt^3$$. The dimensions of $$a$$ and $$d$$ are -
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$$L,T^{-3}$$
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$$L,LT^{-3}$$
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$$L,T^{3}$$
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None of these.
If area $$(A)$$, velocity $$(v)$$ and density $$(\rho)$$ are base units, then the dimensional formula of force can be represented
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$$ \left [ Av\rho \right ] $$
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$$ \left [ Av^{2}\rho \right ] $$
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$$ \left [ Av\rho^{2} \right ] $$
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$$ \left [ A^{2}v\rho \right ] $$
The mass of a bag is $$1.6kg$$. Two objects of mass $$20\,gm$$ and $$15\,gm$$ meausred to an accuracy of $$1gm$$ are added to it. The total mass of the bag calculated to the correct number of significant figures is
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$$1.64\,kg$$
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$$1.635\,kg$$
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$$1.6\,kg$$
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$$1.95\,kg$$
Suppose a quantity x can be dimensionally represented in terms of M, L and $$T$$ that is, $$[x]=M^aL^bT^c$$. The quantity mass.
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Can always be dimensionally represented in terms of L, T and x
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Can never be dimensionally represented in terms of L, T and x
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May be represented in terms of L, T and x if $$a=0$$
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May be represented in terms of L, T and x if $$a\neq 0$$
Using mass (M), length (L), time(T) and current (A) as fundamental quantities,the dimension of permeability is
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$$ [M^{-1} LT^{-2} A] $$
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$$ [M^{-2} LT^{-2} A^{-1}] $$
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$$ [MLT^{-2} A^{-2} ] $$
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$$ [MLT^{-1} A^{-1} ] $$
Taking density $$(\rho)$$, velocity $$(v)$$ and area $$(a)$$ to be fundamental unit then the dimensions of force are:
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$$[av^2\rho]$$
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$$[a^2v\rho]$$
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$$[av\rho^2]$$
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$$[a^0v\rho]$$
A body of mass $$m_{1}$$, moving with uniform velocity of $$40\ m/s$$ collides with another mass $$m_{2}$$ at rest and then the two together begin to move in a uniform velocity of $$30\ m/s$$. The ratio of their masses $$(m_{1}/m_{2})$$ is
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$$0.75$$
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$$1.33$$
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$$3.0$$
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$$4.0$$
Which of the following pairs of physical quantities does not have same dimensional formula?
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Work and torque
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Angular momentum and Plancks constant
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Tension and surface tension
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Impulse and linear momentum
Explanation
a) Work =$$F \times \Delta x=[MLT^{-2}[L]=[ML^2T^{-2}]]$$
Torque=force $$\times$$ distance =$$[ML^2T^{-2}]$$
b) Angular momentum$$mvr=[M][LT^{1}][L]=[ML^2T^{-1}]$$
Plank's constant=$$\dfrac{E}{V}=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$$
c) Tension (force)=$$[MLT^{-2}]$$
Surface tension =$$\dfrac{force}{length}=\dfrac{[MLT^{-2}]}{[L]}=[ML^0T^{-2}]$$
d) Impulse $$=F \times \Delta t=[MLT^{-2}][T]=[MLT^{-1}]$$
Momentum =$$mass \times velocity=[M][LT^{-1}]=[MLT^{-1}]$$
So, among the above pairs only tension and surface tension does not have the same dimensional formula. They both sound similar but they both have different meaning and different applications.
Two forces $$P$$ and $$Q$$ act at a point and have resultant $$R$$. If $$Q$$ is replaced by $$\dfrac{\left ( R^{2}-P^{2} \right )}{Q} $$ acting in the direction opposite to that of $$Q$$, the resultant:
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remains same
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becomes half
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becomes twice
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none of these
The work done by a gas molecule in an isolated system is given by, $$\displaystyle W = \alpha \beta^{2} e^{-\frac{x^{2}}{\alpha kT}}$$, where $$x$$ is the displacement, $$k$$ is the Boltzmann constant and $$T$$ is the temperature, $$\alpha$$ and $$\beta$$ are constants. Then the dimension of $$\beta$$ will be:
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$$[ML^{2} T^{-2}]$$
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$$[MLT^{-2}] $$
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$$[M^{2} LT^{-2}] $$
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$$[M^{0}L T^{0}] $$
Explanation
$$ \dfrac{x^2}{\alpha kT} \rightarrow $$ dimensionless
$$\Rightarrow [\alpha] = \dfrac{[x^2]}{[kT]} = \dfrac{L^2}{ML^2 T^{-2}} = M^{-1} T^{2} $$
Now $$[W] = [\alpha] [\beta]^{2} $$
$$ [\beta] = \displaystyle \sqrt{\dfrac{ML^2 T^{-2}}{M^{-1} T^{2}}} = M^{1} L^{1} T^{-2} $$
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