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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A capacitor `C = (2.0 +- 0.1)muF` is charged to a voltage `V = (20 +- 0.5)` volt. Calculate the charge Q with error limits. |
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Answer» `Q = CV = 2.0xx20 = 40` micro coulomb `=40xx10^(-6)C` `(DeltaQ)/(Q) = (DeltaC)/(C ) +(DeltaV)/(V) = (0.1)/(2.0)+(0.5)20 = (3)/(40)` `DeltaQ = (3)/(40)xxQ = 3.0 muC` `Hence, Q = (40 +- 3.0)xx10^(-6)C` |
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| 102. |
The two specific heat capacities of a gas measured as `C_p = (12.28 +- 0.2)` units and `(C_(upsilon) = (3.97 +-0.3)` units. Find the value of gas constant R. |
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Answer» Here, `C_p = (12.28 +- 0.2)units` `C_(upsilon) = (3.97 +-0.3)units, R = ?` As `R = C_p - C_(upsilon)` `:. R = 12.28 - 3.97 = 8.31units` `DeltaR = +-(DeltaC_p + DeltaC_(upsilon) = +-(0.2 + 0.3) =+-0.5units` `Here, R = (8.31 +- 0.5)units` |
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| 103. |
The specific heats of a gas are measured as `C_p = (12.28 +-0.2)` units and C_(upsilon =(3.97 +- 03)` units. Find the value of gas constant R and percentage error in R. |
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Answer» Correct Answer - `(8.31 +- 0.5) units ; +- 6.016 %` Here, `C_p =(12.28 +- 0.2) units` `C_(upsilon) = (3.97 +- 0.3) units` `R = C_p - C_(upsilon)` `=(12.28 - 3.97) +- (0.2 + 0.3)` `=(8.31 +- 0.5 units` Percentage error in R `=(DeltaR)/(R )xx100 = ((Delta C_p +DeltaC_(upsilon))/(C_p - C_(upsilon)) xx100)` `=((0.2+0.3))/(8.31)xx100 = +-6.016%` |
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| 104. |
The lengths and breadth of a rectangle are `(5.7 +- 0.1)cm and (2.4 +- 0.2)` cm. Calculate area of the rectangle with error limits. |
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Answer» Correct Answer - `(13.7 +- 1.4)sq. cm` Here, `I = (5.7 +- 0.1)cm, b = (2.4 +- 0.2)cm` `A = Ixxb =5.7xx2.4 = 13.68sq. Cm.` = 13.7sq.cm `(DeltaA)/(A) = +- ((DeltaI)/(I) +(Deltab)/(b)) =+-((0.1)/(5.7) +(0.2)/(2.4))` `=+-(0.0175+0.0833)` `DeltaA =+-(0.1008)xx13.7 = +-1.38sq.cm.` `+- 1.4sq. cm` Hence area = (1.37+- 1.4)sq. cm |
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| 105. |
The lengths of the two pieces of wire are `l_1 =(35.2 +- 0.1) cm and l_2 = (47.4 +- 0.2)` cm. what is the total length of wire with error limits ? |
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Answer» Correct Answer - `(82 +- 0.3) cm` Here, `I_1 = (35.2 +- 0.1cm)` `I_2 = (47.4 +- 0.2)cm` `"Total length"` = I_1 + I_2 = (35.2 +47.4)cm` =82.6cm `"Error" =+- (0.1 + 0.2) =+-0.3cm` `:. "Total length" = (82.6 +- 0.03)cm` |
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| 106. |
Express unified atomic mass unit in kg. |
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Answer» Unified atomic mass unit, i.e., ` 1 a.m.u. = 1.67xx10^(-27)kg` Infact, by definiton, `1 a.m.u. = (1)/(12) xx` mass of atom of `C^(12)` `=(1)/(12)xx(12g)/(6.023xx10^(23)) = 1.67x10^(-24)g= 1.67xx10^(-27)kg` |
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| 107. |
A function `f(theta)` is defined as : `f(theta) = 1 - theta+(theta^2)/(2!) - (theta^3)/(3!) +(theta^4)/(4!)….` why is it necessary for `theta` to be a dimensionless quantity ? |
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Answer» `Here, f(theta) = 1 +(theta^2)/(2!) - (theta^3)/(3!)+ (theta^4)/(4!) - …… .` As `f(theta)` is sum of different powers of `theta,` therefore `theta` must be dimensionless. This is because quantities having different dimensions cannot be added. |
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| 108. |
The mass of sun is of the order ofA. `10^(55)kg`B. `10^(42)kg`C. `10^(-30)kg`D. `10^(30)kg` |
| Answer» Correct Answer - (d) | |
| 109. |
The parallax method has been used for measuring distances of stars, which areA. lass then 100 light years awayB. more the 100 light years awayC. neither less nor more then 100 light years awayD. none of above |
| Answer» Correct Answer - (a) | |
| 110. |
Time inerval between two successive heart beats is of the order ofA. 10 sB. `10^(-2)s`C. `10^0s`D. `10^(-1)s` |
| Answer» Correct Answer - (c ) | |
| 111. |
Will five litres of benzene weigh more in summer of winter ? Comment. |
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Answer» Weight = mg = `(V_(rho))`g where `rho` is density. Now, V and g are constant. Therefore weight `prop rho` in summer, temerpature ries. Therefore, due to v thermal expansion density `rho` decreases, and hence, the weight decreases. Thus 5 litres of benzene will weigh more in winter than in summer. |
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| 112. |
Answer the following giving reasons in brief : Is the time variation of position , shown in the figure observed in nature ? |
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Answer» No, such a time varation of position is not observed in nature. This is because (i) At a particular instant of time, an object cannot be present at two different positions, (ii) Time increases first and then decreases indicating backward flow of time, which is not possible. |
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| 113. |
Name three units which can be used for measuring large masses. |
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Answer» 1 slug = 14.57 kg , 1 quintal = 100kg and 1 metric ton= 1000kg |
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| 114. |
What is the average distance of moon from earth ? |
| Answer» The distance of moon from earht is of the order of `10^8m.` | |
| 115. |
Total time taken by a laser beam to return to the earth after reflection from the moon is 2.56 s. Calculate the distance of moon from the earth. Take velocity of light in vacuum `=3xx10^8m//s.` |
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Answer» Correct Answer - `3.84xx10^km` Here, `t = 2.56s, r = ? C=3xx10^8m//s` AS `r = cxx(t)/(2)` `:. r= 3xx10^8xx(2.56)/(2) = 3.84xx10^8m` `=3.84 xx10^5 km. |
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| 116. |
The distance of nearest star from earth is `10^(13)km.` Calculate the time taken by a laser beam to return to earth after reflection from the star. |
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Answer» Here,`x =10^(13) km = 10^(16)m` Velocity of laser beam, `c =3xx10^8m//s` As `x = (cxxt)/(2) :. t= (2x)/(c ) = (2xx10^(16))/(3xx10^8) s = 6.67xx10^7s` |
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| 117. |
why is it convenient to express the distancek of stars in terms of light year rather then in metre or kilometre ? |
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Answer» One light year `= 9.46xx10^(15)m` `=9.46xx10^(12)km.` As the distances of stars are extra- ordinarily large, so it is convenient to express then in terms of light year rathar then in meter or kilometer. |
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| 118. |
Express 1 light year in terms of metre. What is its order of magnitude? |
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Answer» 1 light year = distance travelled by light in vacuum in one year = `cxxt` `=3xx10^8 m//s 365.25xx24xx60xx60s` `= 9.46xx10^(15) metre` `= 0.946 xx10^(16) m` Order of magnitude of light year = 16 |
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| 119. |
Which unit is used to measure size of a nucleus ? |
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Answer» The size of nucleus is measuredc in fermi `1 fermi = 10^(-15)m` |
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| 120. |
Arrange four types of basic forces in the order of increasing strength. |
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Answer» Gravitational forces , weak forces , Electromagnetic forces , Nuclear forces. |
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| 121. |
Masses of atomic`//` subatomic particles are measured using a …………….. . |
| Answer» Correct Answer - mass spectrograph | |
| 122. |
What contribution has Physics made in the development fo biological sciences ? |
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Answer» (i) Electorn microscope using which we can observe the strucutre of a cell. (ii) X-ray and neutron diffrection techniques for studying structure of nucleic acids which control the process of life activity. (iii) Radio - isotopes for radiation treatment of cancer and other disesases. |
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| 123. |
Fill in the blanks (a) The volume of a cube of side 1 cm isk equal to….. `m^3` (b) the surface area fo a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to …. `(mm)^2` (c ) A vehical moving with a speed of `18km h^(-1)` covers ....m in 1s. (d) The relative density of lead is 11.3. its density is ......g `cm^(-3) or .....kg m^(-3)` |
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Answer» (a) Here, length of side, `L = 1 cm =10^(-2)m` Volume of cube `= L^3 = (10^(-2)m)^3 = 10^(-6)m^3` (b) Here, r = 2.0cm = 20mm, h= 10.0cm = 100cm Surface area of solid cylinder ` =(2pi r) h = 2xx(22)/(7)xx20xx100mm^2 = 1.26xx10^4 mm^2` (c ) Here, speed `upsilon = 18 km h^(-1) = (18xx1000m)/(60xx60s) = 5ms^(-1) = 5ms^(-1)` :. Distance covered in 1 second =5 m Here, relative density = 11.3 `:. dentisy = 11.3 g//cc = (11.3xx10^(-3)kg)/((10^(-2)m)^2) = 11.3xx10^3 kg m^(-3)` |
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| 124. |
What is the order of mass fo uiverse. |
| Answer» Mass of universe is of the order of `10^(55)kg.` | |
| 125. |
To measure radus of curvature of a convex mirror unsing a spherometer, it was found that `I = (4.4+-0.1)` cm and h =(0.085 +- 0.001)` cm. Calculate the maximum possible error in the radius of curvature. |
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Answer» Correct Answer - `+-2.5cm` Here, `I =(4.4 +-0.1)cm` `h = (0.085 +- 0.001)cm.` From `R = (I^2)/(6h) +(h)/(2)` `R = ((4.4)^2)/(6xx0.085) +(0.085)/(2)` `=37.96 +0.042 = 38.002 =38cm` (rounded to have two significant figures) `(DeltaR)/(R ) =+-[2((DeltaI)/(l)) +(Deltah)/(h) +(Deltah)/(h)]` `=+-[(2xx0.1)/(4.4) +(0.001)/(0.085) +(0.001)/(0/0.085)]` `(DeltaR)/( r) = +-0.067` `DeltaR = +- 0.067 R= +-0.067x38 = +-2.546` `Delta R= +- 2.5cm` (rounded to have two significant figures) |
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| 126. |
The heat generated in a circuit is given by `Q = I^(2) Rt` , where `I` is current , `R` is resistance , and `t` is time . If the percentage errors in measuring `I , R , and t are 2% , 1% , and 1%` , respectively , then the maximum error in measuring heat will be |
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Answer» Correct Answer - `+- 6%` As `H = (I^2Rt)/(4.2)cal` `:. (DeltaH)/(H) = +-((2DeltaI)/(I) +(DeltaR)/(R )+(Deltat)/(t))` `=+-(2xx2% +1%+1%) =-6%` |
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| 127. |
State the number of significant figures in the following : (i) 0.0070300 m (ii) `2.73xx10^(-4)kg` (iii) 1.0850m (iv) `5.097xx10^3s` |
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Answer» Correct Answer - 5 ; 2, 5 ; 4 (i) 0.0070300m : Five : 7, 0, 3, 0, 0 (ii) `2.73xx10^(-4)kg : Three : 2,7,3` (iii) 1.0850 m: Five : 1,0,8,5,0 (iv) `5.097xx10^3s : Four : 5, 0,9,7` |
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| 128. |
As is known, the result of an experiment is calculated by performing mathematical operation (like addition, subtraction, multiplication, division, etc.) on several measurements, which have different degrees of accuracy. It has been established that (a) When `x= a+b , Delta x = +- (Delta a + Delta b)` (b) When `x= a -b , Delta x = +- (Delta a + Deltab)` (c ) When `x = axxb , (Deltax)/(x) = +- ((Deltaa)/(a) +(Deltab)/(b))` (d) When `x = (a)/(b) , (Deltax)/(x) = +-((Deltaa)/(a) + (Deltab)/(b))` Read the above paragraph and answer the following questions : (i) Why is absolute error in `x = (a -b)`, sun of the absolute error in a and b ? (ii) Why is fractional error in `x=(a)/(b)` , sun of fractional error in a and b ? (iii) What do you learn from this ? |
| Answer» (i) We have made error twice , in measuring a and in measuring b. Therefore, absoute error in x = (a -b ) has to be sum of the two errors. (ii) Same argument applies to fractional error in x = a//b. (iii) From this study, we learn that errors committed any number of times just add. No mathematical operation can reduce the net overall error. So be warned ! Multiple errors are going to cost you dearly. | |
| 129. |
On the basis of dimensional arguments, rule out the wrong relation for Kinetic Energy. (i)`(3)/(16)m upsilon^2` (ii) `(1)/(2)m upsilon^2 + ma` |
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Answer» Correct Answer - Relation (ii) is wrong `K.E. = (1)/(2)mv^2 = M(LT^(-1))^2 = [ML^2T^(-2)]` `Now, (1)/(16)m v^2 = M(LT^(-1))^2 = [ML^2T^(-2)]` :. This formula is dimensionally correct for K.E. (ii) `(1)/(2)m upsilon +ma = M(LT^(-1))^2 +MLT^(-2)` This is not permissibal as quantities with different dimension cannot be added. Hence Relation(ii) is wrong. |
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| 130. |
If velocity of light is taken as the unit of velocity and an year is taken as the unit of time, what is the unit of length? What is it called? |
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Answer» Correct Answer - `9.46xx10^(15)m ; one light year` As velocity ` = (distance)/(time)` :. Distance = velocity xx time `= (3xx10^8ms^(-1))xx(365xx24xx60xx60)s` `=9.46xx10^(15)m` it is called one light year. |
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| 131. |
Out of the formulae y =a sin `2pi t//T` and y = a sin `upsilon t` for the displacement y of particle undergoing a periodic motion, rule out the wrong formula on the basis of dimensions. Symbols have standard meaning. |
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Answer» (i) `y = a sin(2pi t)/(T)` `LHS =y = [L]:` `RHS = a sin(2pi t)/(T) = L sin (T)/(T) = [L]` As dimensions on both sides are same, this formula is correct. (ii) `y = a sin upsilon t` LHS = y = [L], `RHS =a sin upsilon t = L sin (LT^(-1). T) = L sin L` As angle has to be dimensionless, this formula is wrong. |
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| 132. |
The speed of transverse wave `v` in a stretched string depend on length tension `T` in the string and liner mass density (mass per unit length). `mu`. Find the relation using method of dimensions. |
| Answer» Correct Answer - `upsilon = ksqrt(T//m)` | |
| 133. |
The mass of a body is `5xx10^(-6)kg.` What is this mass in `(alpha)` gram (b) milligram (c ) microgram? |
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Answer» Here, `n_1 = 5xx10^(-6), u_1 = 1kg` `n_2 = ? U_2 =1 gram` From `n_2 u_2 =n_1u_1 =n_2 = (n_1u_1)/(u_2) =(5xx10^(-6)xx1kg)/(1 gram) = (5xx10^(-6)kg)/(10^(-3)kg) = 5xx10^(-3)g` (b) `n_2 =? U_2 =1 milligram` `n_2 = (n_1u_1)/(u_2) =(5xx10^(-6)xx1kg)/(1 milligram) =(5xx10^(-6)kg)/(10^(-6)kg) = 5mg` (c ) `n_2 =? u_2 = 1 microgram` `n_2 = (n_1u_1)/(u_2) = (5xx10^(-6)xx1kg)/(1 microgram) = (5xx10^(-6)kg)/(10^(-9)kg) = 10^3 microgram` |
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| 134. |
The dimensional formula for permeability of free space, `mu_(0)` isA. `[MLT^(-2)A^(-2)]`B. `[ML^(-1)T^(2)A^(-2)]`C. `[ML^(-1)T^(-2)A^(2)]`D. `[MLT^(-2)A^(-1)]` |
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Answer» Correct Answer - A From Biot-Savart law `dB - (mu_(0))/(4pi) (Idlsin theta)/(r^(2))` `Idl =` current element `r =` displacement vector `mu_(0) = (4pi^(2)(dB))/(Idl sin theta) =([L^(2)][MT^(-2)A^(-1)])/([A][L]) = [MLT^(-2)A^(-2)]` |
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| 135. |
A certain body weighs `22.42 g` and has a measured volumen of `4.7 c c`. The possible error in the measurement of mass and volumen are `0.01g` and `0.1 c c`. Then, maximum error in the density will beA. `22%`B. `2%`C. `0.2%`D. `0.02%` |
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Answer» Correct Answer - B Density `= ("Mass")/("Volume")` `rho = (m)/(V)` `:. (Deltarho)/(rho) = (Deltam)/(m) + (DeltaV)/(V)` Here, `Deltam = 0.01m, 22.42` `DeltaV = 0.1V, 4.7` `:. (Deltarho)/(rho) = ((0.01)/(22.42)+(0.1)/(4.7)) xx 100 = 2%` |
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| 136. |
The dimensional formula for magnetic flux isA. `[ML^(2)T^(-2)A^(-1)]`B. `[ML^(3)T^(-2)A^(-2)]`C. `[M^(0)L^(-2)T^(2)A^(-2)]`D. `[ML^(2)T^(-1)A^(2)]` |
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Answer» Correct Answer - A Mathematically, magnetic flux `phi = B A` ...(i) but magnetic force `F = Bil` or `B = (F)/(il)` Putting the value of B in Eq.(i), we have `phi = (F)/(il)A` Thus, dimensions of `phi = ([MLT^(-2)][L^(2)])/([AL])` `= [ML^(2)T^(-2)A^(-1)]` |
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| 137. |
The error in the measurement of radius of a sphere in `+-4%` What would be the error in volume of the sphere? |
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Answer» Correct Answer - `+-12%` Here, `(Deltar)/(r ) = +-4%` As volume `V = (4)/(3) pi r^3` `:. (DeltaV)/(V) =+-3((Deltar)/(r )) = +-3xx4% = +-12%` |
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| 138. |
An equation is given as `(p +(a)/(V^(2))) = b(theta)/(V)`,where `p=` pressure `V=` volumen and `theta =` absolute temperature. If a and b are constants, then dimensions of a will beA. `[ML^(5)T^(-2)]`B. `[M^(-1)L^(5)T^(-2)]`C. `[ML^(-5)T^(-1)]`D. `[ML^(5)T]` |
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Answer» Correct Answer - A From principle of homogeneity of dimensions. Dimensions of `p =` dimensions of `(a)/(V^(2))` `p = (a)/(V^(2)) rArr a = pV^(2)` `=[ML^(-1)T^(-2)] [L^(3)]^(2) = [ML^(5)T^(-2)]` |
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| 139. |
Which of the following will have the dimensions of time ?A. `LC`B. `(R)/(L)`C. `(L)/(R)`D. `(C)/(L)` |
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Answer» Correct Answer - C `(L)/(R)` is time constant of R-L circuit so, dimensions of `(L)/(R)` is same as that of time. ALternative `("Dimensions of"L)/("Dimensions of "R) = [T]` |
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| 140. |
The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of density will beA. `7%`B. `9%`C. `12%`D. `13%` |
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Answer» Correct Answer - D As density `rho = (m)/(V) = (m)/(I^(3))` `:. (Delta rho)/(rho) xx 100 = +- ((Deltam)/(m) +3 (Deltal)/(l)) xx 100%` `= +- (4+3 xx 3) = +- 13%` |
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| 141. |
The mass of a proton is `1.67xx10^(-27)kg.` How may protons would make 1 gram? |
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Answer» Here `m_(p) = 1.67xx10^(-27)kg`, `"total mass", m =1g = 10^(-3)kg` `:. "Number of protons"=m/m_(p)=(10^(-3Kg))/(1.67xx10^(-27)Kg)` `=5.99xx10^(23)` |
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| 142. |
An area of ………….. `m^2` is called ……………… . |
| Answer» Correct Answer - `10^4` ; one hectare. | |
| 143. |
The radius of proton in of the order ofA. `10^(15)m`B. `10^(-15)m`C. `10^(-14)m`D. `10^(-31)m` |
| Answer» Correct Answer - (b) | |
| 144. |
The order of magnitude of height of man isA. zeroB. 1C. -1D. none of these |
| Answer» Correct Answer - (a) | |
| 145. |
What is the order of magnitude of 499 and 0.050 ? |
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Answer» `499 = 4.99 xx10^2 = 1 xx10^2 ,` order of magnitude is 2 `0.050 = 5.0 xx 10^(-2) rarr 10xx10^(-2) = 10^9-1 ,` order of magnitude is - 1. |
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| 146. |
What is the order of magnitude of radius of earth ? |
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Answer» `R = 6400 km = 6.4 xx 10^6 m ~= 10xx10^6 m` `= 10^7m` Order of magnitude of R is 7. |
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| 147. |
A man walking briskly in rain with speed `upsilon` must slant his umbrella forward making an angle(with the vertical). A student derives the following relation between `theta and upsilonn : tan theta = upsilon` and checks that the relation has a correct limits : as `upsilon rarr 0, theta rarr 0.` as expected. (we are assuming there is no strong wind and that the rain vertically for a stationary man). Do you think this relation can be correct ? if not, guess the correct relation. |
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Answer» The relation `tan theta = upsilon has a correct limit, as upsilon rarr 0 , theta rarr 0.` However, `RHS = tan theta = [M^0 L^0 T] and LHS = upsilon = [M^0 L^1 T^(-1)].` Therefore, the relation is not correct dimensionally. As we go through unit 3 of the book, we shall find that the correct relation is tan `theta = (upsilon^2)/(rg)` |
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| 148. |
A book with many printing errors contains four different forumlae for the displacement y of a particle undergoing a certain periodic motion : (i) `y = a sin (2pi t)/(T)` (ii) `y = a sin upsilon t` (iii) `y = (a)/(T) sin (t)/(a)` (iv) `y= (a)/(sqrt2)[sin(2pi t)/(T) + cos (2pi t)/(T)]` Here, a is maximum displacement of particle, `upsilon` is speed of particle, T is time period of motion. Rule out the wrong forumlae on dimensinal grounds. |
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Answer» The argument of a trigonmietrical funciton, i.e., angle is dimensionless. Now, in (i) `(2pi t)/(T) = (T)/(T) = 1 =(M^0 L^0 T^0)` …. "Dimensionless"` (ii) `upsilon t = (LT^(-1))(T) = L = [M^0 L^1 T^0] …." Not dimensionless"` (iii) `(t)/(a) = (T)/(L) = [L^(-1) T^1] ....." not dimensionless"` (iv) `(2pit)/(T) = (T)/(T) = 1 = [M^0L^0 T^0] ...."dimensionless"` :. formulea (ii) and (iii) are wrong. |
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| 149. |
Time taken by a body in `(20 +-0.2)` second in undergoing a displacement of `(200 +-5)`m. Calculate the percentage error in calculation of velcoity. |
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Answer» Correct Answer - `+- 3.5%` `upsilon =(s)/(t) = (200)/(20) =10m//s` `(Deltaupsilon)/(upsilon)xx100 = +-((Deltas)/(s) +(Detlat)/(t))xx100` `=+-((5)/(200) +(0.2)/(20))xx100` `=+- (2.5+1) =+-3.5%` |
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| 150. |
The length breadth and thickness of a metal sheet are 4.234 m, 1.005 m and 2.01 cm respectively. Given the area and volume of the sheet to correct number of significant figure. |
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Answer» Here, length I = 4.234 m, breadth b = 1.005m, thickness, `t = 2.01 cm = 2.01xx10^(-2)m` Area of the sheet = `2 (lxxb + bxxt + +txxl) = 2(4.234xx1.005 + 1.005 xx0.0201 + 0.0201 xx 4.234)` `= 2 (4.3604739) = 8.7209478 m^2` As area can contain a maximum of three significant digits, therefore, rounding off, we get `Area = 8.72 m^2` Also, volume = `lxxbxxt` `V = 4.234xx1.005xx0.0201 = 0.0855289 = 0.0855 m^3` ("containing three singinficant figures")` |
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