Q # 1. Name several repetitive phenomenon occurring in nature which can serve as reasonable time standards.
Ans. Any natural phenomenon that repeats itself after exactly same time interval can be used as time standard. The following natural phenomenon can be used as time standard.
- The rotation of earth around the sun and about its own axis
- The rotation of moon around earth
- Atomic vibrations in solids
Q # 2. Give the drawbacks to use the time period of a pendulum as a time standard.
Ans. The time period of the simple pendulum depends upon its length l and value of g (gravitational acceleration) at any place. Therefore, the drawbacks to use the time period of a pendulum as a time standard are
- The value of g changes at different places
- The variation in the length of simple pendulum due to change in temperature in different seasons
- Air resistance may affect the time period of simple pendulum
Q # 3. Why we use it useful to have two units for the amount of substance, the kilogram and the mole?
Ans. The kilogram and mole are the units to determine the amount of a substance. Both units are useful in different cases describe below
- The unit kilogram is useful when we want to desçibe the macroscopic behavior of an object without considering the interaction between the atoms,present in it
- The unit mole is useful when we want consider a fix number of atoms of a system. It is used to determine the microscopic behavior of any object.
Q # 4. Three students measured the length of a needle with a scale on which minimum division is 1 mm and recorded as (i) 0.2145 m (ii) 0.21 m (iii) 0.214 m. Which record is correct and why?
Ans. The record (iii) is correct.
Reason: As the scale used for measurement has the least count of 1 mm=0.001 m. So the reading must be taken up to three decimal places. Therefore, the reading 0.214 m is correct.
Q # 5. An old saying is that "A chain is only as strong as its weakest link". What analogous statement can you make regarding experimental data used in computation?
Ans. The analogous statement for experimental data used in computation will be
"A result obtained by mathematical computation of experimental data is only as much accurate as its least accurate reading in measurements".
Q # 6. The time period of the simple pendulum is measured by a stop watch. What types of errors are possible in the time period?
Ans. The possible errors that might occur are the personal error and systematic error. The personal error occurs due to negligence or inexperience of a person, while the systematic error may be due to the poor calibration of equipment or incorrect marking etc.
Q # 7. Does the dimensional analysis gives any information on constant of proportionality that may appear in an algebraic expression. Explain?
Ans. Dimension analysis does not give any information about constant of proportionality in any expression. This constant can be determined experimentally of theoretically.
Example: In the expression of time period of simple pendulum, the constant of proportionality cannot be determined from dimension analysis.
Q # 8. What are the dimensions of (i) Pressure (ii) Density
Ans (i) By definition,
$${\text {Pressure}}=\frac{\text { Force }}{\text { Area }}$$
$${\text { Dimension of Pressure}} =\frac{\text { Dimensions of Force }}{\text { Dimensions of Area }} $$
$$\Rightarrow[P]=\frac{[F]}{[\mathrm{A}]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}$$
$$\Rightarrow[P]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$$
(ii) By definition,
$${\text {Density}}=\frac{\text { Mass }}{\text { Volume}}$$
$${\text {Dimension of Density}}=\frac{\text { Dimensions of Mass }}{\text { Dimensions of Volume }}$$
$$\Rightarrow[\rho]=\frac{[\text { mass }]}{[\text { volume }]}$$
$$\Rightarrow[\rho]=\left[\mathrm{ML}^{-3}\right]$$
Q # 9. The wavelength $\lambda$ of a wave depends on the speed $v$ of the wave and its frequency $f$. Decide which of the following is correct, $f=v \lambda \quad$ or (ii) $f=\frac{v}{\lambda}$?
(i) $f=v \lambda$
Dimension of LHS $=[\mathrm{f}]=\left[\mathrm{T}^{-1}\right]$
Dimension of RHS $=[v \lambda]=[v][\lambda]$
$$\begin{aligned}& \because[\mathrm{v}]=\left[\mathrm{LT}^{-1}\right] \\& \because[\lambda]=[\mathrm{L}]\end{aligned}$$
Dimension of RHS $=[\mathrm{v} \lambda]=\left[\mathrm{LT}^{-1}\right][\mathrm{L}]=\left[\mathrm{L}^{2} \mathrm{~T}^{-1}\right]$
As Dimension of LHS $\neq$ Dimension of RHS
Hence, the equation $f=v \lambda$ is not dimensionally correct.
(ii) $\mathrm{f}=\frac{\mathrm{v}}{\lambda}$
Dimension of LHS $=[\mathrm{f}]=\left[\mathrm{T}^{-1}\right]$
Dimension of RHS $=\left[\frac{\mathrm{v}}{\lambda}\right]=\frac{[\mathrm{v}]}{[\lambda]}=\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{L}]}=\left[\mathrm{T}^{-1}\right]$
As
Dimension of LHS = Dimension of RHS
Hence, the equation $\mathbf{f}=\frac{\mathbf{v}}{\lambda}$ is dimensionally correct.
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