CBSE-XI-Physics
01: Introduction to Physics
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- #Section : B
- Qstn #1Which of the following sets cannot enter into the list of fundamental quantities in any system of units?
-(a) length, mass and velocity,
(b) length, time and velocity,
(c) mass, time and velocity,
(d) length, time and mass.digAnsr: bAns : (b) length, time and velocity
We define length and time separately as it is not possible to define velocity without using these quantities. This means that onefundamental quantity depends on the other. So, these quantities cannot be listed as fundamental quantities in any system of units.
- Qstn #2A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numerical value. If the result is expressed in various units then
-(a)
n∞sizeofu
(b)
n∞u2
(c)
n∞u
(d)
n∞1u.digAnsr: dAns : (d) `` n\propto \frac{1}{u}``
The larger the unit used to express the physical quantity, the lesser will be the numerical value.
Example:
1 kg of sugar can be expressed as 1000 g or 1000000 mg of sugar.
Here, g (gram) is the larger quantity as compared to mg (milligram), but the numerical value used with gram is lesser than the numerical value used with milligram.
- Qstn #3Suppose a quantity x can be dimensionally represented in terms of M, L and T, that is,
[x]=``M^aL^bT^c``. The quantity mass
-(a) can always be dimensionally represented in terms of L, T and x,
(b) can never be dimensionally represented in terms of L, T and x,
(c) may be represented in terms of L, T and x if a = 0,
(d) may be represented in terms of L, T and x if
a≠0.digAnsr: dAns : (d) may be represented in terms of L, T and x if `` a\ne 0``
If a = 0, then we cannot represent mass dimensionally in terms of L, T and x, otherwise it can be represented in terms of L, T and x.
- Qstn #4A dimensionless quantity
-(a) never has a unit,
(b) always has a unit,
(c) may have a unit,
(d) does not exist.digAnsr: aAns : (a) may have a unit
Dimensionless quantities may have units.
- Qstn #5A unitless quantity
-(a) never has a non-zero dimension,
(b) always has a non-zero dimension,
(c) may have a non-zero dimension,
(d) does not exist.digAnsr: aAns : (a) never has a non-zero dimension
A unitless quantity never has a non-zero dimension.
- Qstn #6∫dx2ax-x2=ansin-1xa-1.
The value of n is
-(a) 0
(b) -1
(c) 1
(d) none of these.
You may use dimensional analysis to solve the problem.digAnsr: aAns : (a) 0
[ax] = [x2]
⇒ [a] = [x] ...(1)
Dimension of LHS = Dimension of RHS
`` \Rightarrow \left[\frac{dx}{\sqrt{{x}^{2}}}\right]=\left[{a}^{n}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \left[\frac{L}{L}\right]=\left[{a}^{n}\right]...\left(2\right)\phantom{\rule{0ex}{0ex}}\left[{L}^{0}\right]=\left[{a}^{n}\right]\phantom{\rule{0ex}{0ex}}n=0\phantom{\rule{0ex}{0ex}}``
- #Section : C
- Qstn #1The dimensions ML-1 T-2 may correspond to
-(a) work done by a force
(b) linear momentum
(c) pressure
(d) energy per unit volume.digAnsr: c,dAns : (c) pressure
(d) energy per unit volume
[Work done] = [ML2T-2]
[Linear momentum] = [MLT-1]
[Pressure] = [ML-1 T-2]
[Energy per unit volume] = [ML-1 T-2]
From the above, we can see that pressure and energy per unit volume have the same dimension, i.e., ML-1 T-2.
- Qstn #2Choose the correct statements(s):
-(a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(c) A dimensionally incorrect equation may be correct.
(d) A dimensionally incorrect equation may be incorrect.digAnsr: a,b,dAns : (a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(d) A dimensionally incorrect equation may be incorrect.
It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible.
- Qstn #3Choose the correct statements(s):
-(a) All quantities may be represented dimensionally in terms of the base quantities.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities.
(c) The dimensions of a base quantity in other base quantities is always zero.
(d) The dimension of a derived quantity is never zero in any base quantity.digAnsr: a,b,cAns : The statements which are correct are:
(a) All quantities may be represented dimensionally in terms of the base quantities.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities.
(c) The dimensions of a base quantity in other base quantities is always zero.
Statement (d) is not correct because A derived quantity can exist which is dimensionless for example fine structure constant which is given by
`` \alpha =\frac{2\pi {e}^{2}}{hc}=\frac{1}{137} \phantom{\rule{0ex}{0ex}}\, \mathrm{where} \, \mathrm{e}\, \mathrm{is}\, \mathrm{the}\, \mathrm{electric}\, \mathrm{charge}\, \mathrm{and}\, \mathrm{c}\, \mathrm{is}\, \mathrm{the}\, \mathrm{speed}\, \mathrm{of}\, \mathrm{light}\, \mathrm{and}\, \mathrm{h}\, \mathrm{is}\, \mathrm{Planks}\, \mathrm{constant}.\phantom{\rule{0ex}{0ex}}\, \mathrm{\alpha }\, \mathrm{is}\, \mathrm{a}\, \mathrm{derived}\, \mathrm{quantity}\, \mathrm{and}\, \mathrm{is}\, \mathrm{dimensionless}.``
- #Section : D
- #1-alinear momentum,Ans : Linear momentum = mv
Here, [m] = [M] and [v] = [``LT^{-1}``]
∴ Dimension of linear momentum, [mv] = [``MLT^{-1}``]
- #1-bfrequency andAns : Frequency = `` \frac{1}{\, \mathrm{Time}}``
∴ Dimension of frequence = `` \left[\frac{1}{T}\right]=\left[{\, \mathrm{M}}^{0}{\, \mathrm{L}}^{0}{\, \mathrm{T}}^{-1}\right]``