InterviewSolution
Saved Bookmarks
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.
| 201. |
In the difference of two quantitesA. maximum absolute error is equal to sum of absolute errors in individual quantitesB. maximum absoulte error is equal to difference in absolute errors in individual quantitiesC. Either (a) or (b)D. neither (a) nor (b) |
| Answer» Correct Answer - (a) | |
| 202. |
Chack the correctness of the relations. (i) escape velocity, `upsilon = sqrt((2GM)/(R ))` (ii) `v =(1)/(2l) sqrt((T)/(m))`, where l is length, T is tension and m is mass per unit length of the string. |
|
Answer» Correct Answer - (i) correct (ii) correct (i) `upsilon = sqrt((2GM)/(R ))` `LHS = upsilon = [M^0L^1T^(-1)]` `RHS = sqrt((2GM)/(R )) = sqrt((M^(-1)L^3T^(-2)(M))/(L))` `= sqrtL^2T^(-2) =[M^0L^1T^(-1)]` AS LHS = RHS, dimensionally, therefore the formula is correct. (ii) `v =(1)/(2I)sqrt((T)/(m))` `LHS = v=[M^0L^0T^(-1)]` `RHS = (1)/(2I) sqrt((T)/(m)) = (1)/(L) sqrt((MLT^(-2))/(ML^(-1))` `=(1)/(L)sqrt(L^2T^(-2)) = T^(-1) = [M^0L^0T^(-1)]` As LHS = RHS, dimensionally, therefore the formula is correct. |
|
| 203. |
The prefix femto stands forA. `10^(15)`B. `10^(-15)`C. `10^5`D. `10^(-5)` |
| Answer» Correct Answer - (b) | |
| 204. |
Waber is derived unit ofA. magnetic momentB. luminous fluxC. magnetic fluxD. none of these |
| Answer» Correct Answer - (c ) | |
| 205. |
Which of the following relations is not correct ?A. `1 A.U = 1.496xx 10^(11)m`B. `1ly = 9.46xx10^(15)m`C. `1parsec = 3.084xx10^(16)m`D. `1 ly = 6.3xx10^(-4) A.U. |
| Answer» Correct Answer - (d) | |
| 206. |
Assertion : if the units of force and length are doubled, the unit of energy will be 4 times. Reason : The unit of energy is independent of the units of force and lengthA. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both , Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
|
Answer» Correct Answer - (c ) Here, assertion is true, but reason is false. As Energy = Work = `F xx S` :. When the unit of force and length are doubled the unit of energy will becomes 4 times. Also, unit of energy depends upon force and length. |
|
| 207. |
A standard metre is equal to k wavelengths in vacuum, of the radiation from Krypton 86, where k isA. 165076.37B. 16507637.3C. 1650763.73D. none of these |
| Answer» Correct Answer - (c ) | |
| 208. |
Light emitted by Krypton 86 is 6057.8021 Å. Calculate number of wavelengths of Krypton 86 in one metre. What is the order of magnitude ? |
|
Answer» Here, `lambda= 6057.8021 Å` `=6.0578021xx10^(-7)m` `1m = (1)/(6.0578021xx10^(-7))` wavelengths = `1650, 763.3` wavelengths `1m =1.6507633xx10^6` As `1.65 lt 5`, therefore, order of magnitude = 6. |
|
| 209. |
Statement-1 If error in measurement of distance and time are 3% and 2% respectively, error in calculation of velocity is 5% Statement-2 : Velocity `= (distance)/(time)`A. Statement -1 is true, Statement -2 is true , and Statement -2 is correct explanation of Statement -1.B. Statement -1 is true , Statement -2 is true, but Statement -2 is not a correct explanation of Statement -1.C. Statement-1 is true, but Statement -2 is false.D. Statement-1 is false, but Statement -2 is true. |
|
Answer» Correct Answer - (b) Both the statements are true. But statement -2 is not a correct explanation of statement -1. Infact `upsilon = (L)/(T)` `:. (Delta upsilon)/(upsilon) = +-((Delta L)/(L) + (Delta T)/(T)) = +- (3% + 2%)` `= +- 5%` |
|
| 210. |
Derive an expression for time period (t) of a simple pendulem, which may depend upon : mass of bob (m), length of pendulum (I) and acceleration due to gravity(g). |
|
Answer» Let `t prop m^a l^b g^c` where a, b, c are the dimensions. Or `t = k m^a l^b g^c …(4)` where k is dimensionless constant of proportionality. Writing the dimensions in terms of M, L, T on either side of (4), we get `[M^0 L^0 T^1] = M^a L^b (LT^(-2))^c = M^a L^(b + c) T^(-2c)` Applying the principle of homogeneity of dimensions, we get `a = 0 b+c =0 ....(5)` ` -2 c= 1 or c = -(1)/(2)` From (5),` b=- c = -(-(1)/(2)) = (1)/(2)` `"Putting the value of a,b,c, in"` (4), `"we get"` `t = km^0 I^(1//2) g^(-1//2)` ` t = ksqrt(I)/(g)` Using other methoud, we calculate the value of dimensionless constant, ` k =2pi :, t = 2 pi sqrt(I//g)` |
|
| 211. |
Who first gave the concept of antiparticle ? |
| Answer» Correct Answer - Paul Dirac. | |
| 212. |
A substance weighing 5.74 g occupies a volume of `1.2cm^3.` Caluclate its density with due regard to significant digits. |
|
Answer» `rho = (M)/(V) = (5.74)/(1.2) = 4.78 g//cc = 4.8 g//cc,` rounded off to have two significant digits. |
|
| 213. |
The numbers 2.745 and 2.735 on rounding off to 3 significant figures will giveA. 2.75 and 2.74B. 2.74 and 2.73C. 2.75 and 2.73D. 2.74 and 2.74 |
|
Answer» Correct Answer - (d) On rounding off to 3 significant figures, `2.744 rar 2.74`, as per rules , `2.735 rarr 2.74`, as per rules. |
|
| 214. |
When planet Jupiter is at a distance of 824.7 million km from earth, its angular diameter is measured to be 35.72' of arc. Calculate the diameter of Jupiter. |
|
Answer» Correct Answer - `1.428xx10^5 km` Here, r 824.7 million km `=824.7xx10^6km` `theta = 35.72' = (35.72)/(60xx60)xx(pi)/(180) radian` I = ? From `I =r theta` `I 824.7xx10^6xx(35x72)/(60xx60)xx(pi)/(180)km` `=1.428xx10^5km` |
|
| 215. |
when the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is mieaured to be 35.72' of arc. Calculate the diameter of Jupiter ? |
|
Answer» Here, `r = 824.7xx10^6km.` `theta = 35.72' = (35.72)/(60xx60)xx(pi)/(180)radian` Diameter, I = ? As `I = r theta :. I = 824.7xx10^6xx(35.72xxpi)/(60xx60xx180)km = 1.492xx10^5km` |
|
| 216. |
Given names of a scalar quantity and a vector quantity which have same dimensions. |
|
Answer» Work is scalar and torque is vector. Both have the same dimensions `[ML^2 T^(-2)].` |
|
| 217. |
What is the range of nuclear forces ? |
| Answer» The range of nuclear forces is `~~ 10` fermi. | |
| 218. |
Mechanical energy is always constant. Is the statement ture ? |
|
Answer» No, the statement is true only when internal forces involved are conservation and external forces do no work. |
|
| 219. |
Why do we call Physis an exact science? |
|
Answer» Physcial quantities in Physcies are measured with high precision and accuracy. That is why Physics is called an exact science. |
|
| 220. |
Express 1 micron in metre. |
| Answer» ` 1 micron = 10^(-6)metre.` | |
| 221. |
The moon subtends an angle of 57 minutes at the base line equal to radius of earth. What is the distance of moon from earth. Given radius of earth is 6400 km. |
|
Answer» Here, `theta = 57 min. = (57^@)/(60) = (19)/(20)xx(pi)/(180)rad` I = radius of earth ` =6400km = 6.4xx10^6m = r = ?` From `I =r theta, r= (1)/(theta) = (6.4xx10^6x20xx180)/(19xxpi) m = 3.86xx10^8m` |
|
| 222. |
Should a scientific discovery which has nothing but dangerous consequences for manking be made public? |
|
Answer» Yes, and discovery good or bad, must be made public. Something which appears dangerous today, may be put to use in some other form later. So, a discovery, which reveals a truth of nature. should not be concealed. |
|
| 223. |
Assertion : Light year and year, both measure time. Reason : Because light year is the time light takes to reach the earth from the sun.A. If both Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both , Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
|
Answer» Correct Answer - (d) Light year is unit of distance and not of time. Therefore, both the assertion and reason are false. |
|
| 224. |
Express in scientific notation : (i) 13780 kg (ii) 0.00000523 s |
| Answer» (i) `1.3780 xx 10^4 kg` (ii) `5.23 xx 10^(-6)s` | |
| 225. |
Express the average distance of earth from the sun in (i) light year (ii) per sec. |
| Answer» 1 Astronomical unit, which is equal to `1.496xx10^(11)m.` | |
| 226. |
Average life of a human being is of the order ofA. `10^7s`B. `10^(-7)s`C. `10^9s`D. `10^(-9)s` |
| Answer» Correct Answer - (c ) | |
| 227. |
Quartz crystal clocks have an accuracy of 1 sec in everyA. `10^(-9)s`B. `10^9s`C. `10^(-13)s`D. `10^(13)s` |
| Answer» Correct Answer - (b) | |
| 228. |
Light from the sun reaches the earth approximately in `x xx 10^2` sec, where x is : |
|
Answer» Correct Answer - 5 `t = (s)/(c ) = (1.5xx10^(11)(m))/(3xx10^8(m//s)) = 0.5xx10^3 s` `=5xx10^2s.` :. X =5 |
|
| 229. |
Choose the quatity whose unit is not treated as a fundamental unit.A. lengthB. velocityC. massD. time |
| Answer» Correct Answer - (b) | |
| 230. |
A bigger unit is contained ………………….. Of times in the quantity. |
| Answer» Correct Answer - smaller number | |
| 231. |
Radioactive dating is used for measuring long time intervals of the order ofA. `10^(17)s`B. `10^7s`C. `10^(17)years`D. `10^7 years` |
| Answer» Correct Answer - (a) | |
| 232. |
The process of measurement is basicallyA. a process of comparisonB. a proces of estimationC. a process of easeD. none of these |
| Answer» Correct Answer - (a) | |
| 233. |
The process of measurement is basically a process of comparison. The chosen standard of measurement of a quantity, which has essentially the same nature as that of the quantity is called unit of the quantity. Magnitude of a quantity (Q) = size of its unit `(u)xx` number of times (n) this unit is contained in the quantity. i.e. Q = nu As magnitude of a quantity remains the same, whatever be its units of measurement, therefore, `Q = n_1u_1 = n_2 u_2` Read the above passage and answer the following questions : (i) The value of acceleration due to gravity `(g) = 9.8 m//s^2` How do you express it in `km//hr^2` ? (ii) Our world is a game of numbers. Do you agree ? Justify. |
|
Answer» (i) `g=(9.8m)/(1s)^2 = (9.8xx10^(-3)km)/((1)/(60xx60)hr)^2 = 1.27xx10^5 km//hr^2` (ii) Yes, ceerainly. In every walk of life, we play with number. For example : who is taller, who is havier, who has broken world record in any game : which is the tallest building, which is the highest mountain peak , who is suffering from disbetes, hypertension, liver ailment, kidney ailment and so on. Everything is every sphere of life, revolves around numbers. |
|
| 234. |
In a particular system the units of length, mass and time are chosen to be `10 cm, 10 g` and `0.1 s` respectively. The unit of force in this system will be equal toA. `0.1N`B. `1N`C. `10N`D. `100N` |
|
Answer» Correct Answer - A Force `F = [MLT^(-2)] = (10g) (10cm) (0.1s)^(-2)` Changing these units into MKS system `F = (10^(-2) kg) (10^(-1)m) (10^(-1)s)^(-2)` `10^(-1) N = 0.1N` |
|
| 235. |
The radius of a sphere is measured to be `(2.1+-0.5)` cm. Calculate its surface area with error limits . |
|
Answer» Correct Answer - `(55.4 +- 26.4)cm^2` surface area of a sphere `=4pir^2` |
|
| 236. |
In case of venus, the angle of maximum elongation is found to be approximately `47^@.` Determine the distance between venus and sun `(r_(upsilon e).` and the distance between venus and earth `(r_(upsilon e).` |
|
Answer» Here, angle of maximum elongation `in = 47^@, r_(upsilon s) = ?, r_(upsilon e) = ?` We known, `r_(se) = 1A. U = 1.496xx10^(11)m.` `r_(upsilon s) = r_(es) xx sin in` `=1.496xx 10^(11) sin 47^@` `1.496xx10^(11)xx0.73` ` = 1.09 xx10^(11)m` `r_(upsilon e) =r_(se)xx cos in` `= 1.496xx10^(11) cos 47^@` `=1.496xx10^(11) xx0.68m` `= 1.0xx10^(11)m` |
|
| 237. |
The depth x to which a bullet penetrates a human body depends on (i) coeffeicint of elasticity, `eta` and (ii) KE `(E_k)` of the bullet, By the method of dimensions, show that `x prop ((E_k)/(eta))^(1//3)` |
|
Answer» Let `x prop E_k^a eta^b,` where, a, b are the dimensions `x = K E_k^a eta^b ….(i)` where K is dimansionless constant fo proportionality. Writing the dimensions in (i), we get `[M^0L^1 T^0] = (ML^2 T^(-2))^a (ML^(-1) T^(-2))^b` `=M^(a +b) L^(2a -b) T^(-2a -2b)` Applying the principle of homogeneity of dimensions, we get ` a +b =0 .....(ii)` `2a - b =1 ....(iii)` `-2a -2b =0 ....(iv)` Add (ii) and (iii) : `3a =1, a = 1//3` Form (ii), `b =-a - (1)/(3)`. From (i),` x =K((E_k)/(eta))^(1//3)` Henc, `x prop((E_k)/(eta))^(1//3)` |
|
| 238. |
A large fluid star oscillates in shape under the influence of its own gravitational field. Using dimensional analysis, find the expression for period of oscillation (T) in terms of radius of star (R ), mean density of fluid `(rho)` and universal gravitational constant (G). |
|
Answer» Let `T = KR^a rho^b G^c ….. 9i)` `[M^0 L^0 T^1] = [L]^a [ML^(-3)]^b[M^(-1) L^3 T^(-2)]^c = M^(b-c) L^(a - 3b+ 3c) T^(-2c)` Applying principle of homogeneity of dimensions, we get `b -c =0 , a - 3b + 3c = 0 ,-2c =1, c = -(1)/(2)` `:. b = c = (1)/(2) and a-3(-(1)/(2)) +3 (-(1)/(2)) = 0, a= 0` Putting in (i), we get `T= KR^0 rho^(-1//2) G^(-1//2) = K(rhoG)^(-1//2)` |
|
| 239. |
To study the flow of a liquied through a narrow tube, the following formula is used :` eta = (pi)/(8) (Pr^4)/(VI),` where letters have their usual meaning. The value of P, r, V and I are 76 cm of Hg col. `0.28 cm, 1.2 cm^3 s^(-1) and 18.2cm` respectively. If these quantities are measured to the accuracy of `0.1 cm of Hg col., 0.01 cm, 0.1 cm^3 s^(-1)` and 0.1 cm respectively, find the percentage error in the value of `eta.` |
|
Answer» Here, ` P =76cm` of `Hg col. DeltaP = 0.1 cm` ` r= 0.28 cm Delta = 0.01 cm` `V = 1.2 cm^3 s^(-1) DeltaV = 0.1 cm^3 s^(-1)` `I =18.2 cm DeltaI = 0.1 cm` From `eta = (pi)/(8) (Pr^4)/(VI)` `(Delta eta)/(eta) = (DeltaP)/(P) + 4(Deltar)/(r ) +(DeltaV)/(V) + (DeltaI)/(I) = (0.1)/(76) + (4(0.01))/(0.28) +(0.1)/(1.2) +(0.1)/(18.2)` ` = 0.0013 + 0.1428 + 0.0833 + 0.0054 = 0.2328` `:. (Delta eta)/(eta)xx100 = 0.2328xx100 = 23.28%` |
|
| 240. |
The rate flow (V) of a liquid through a pipe of radius (r ) under a pressure gradient (P//I) is given by `V = (pi)/(8)(P R^4)/(I eta),` Where `eta` is coefficient of visocity of the liquied. Check whether the formula is correct or not. |
|
Answer» `V = volume//sec =[L^3 T^(-1)]` `(P)/(I) = "pressure gradient" = (ML^(-1) T^(-2))/(L)` ` = [ML^(-2)T^(-2)]` Now `L.H.S. = V = [L^3 T^(-1)]` `R.H.S = (pi)/(8)xx(P)/(I)xx(r^4)/(eta) = ((ML^(-2)T^(-2))L^4)/((ML^(-1)T^(-1))) = [L^3 T^(-1)]` As L.H.S. = R.H.S., dimensionly, therefore the formula is correct. |
|
| 241. |
Surface tension of mercury is 540 `dy"ne"//cm`. What will be its value when unit of mass of 1kg. Unit of length is 1m and unit of time is 1 minute? |
|
Answer» Here, `n_1 =540 "dyne"//cm`. `M_1 =1g,L_1 = 1cm, T_1 =1` sec `n_2 =? M_2 =1 kg, L_2 =1m, T_2 = 1 min =60s` Surface tension ` = ("force")/("length") =(MLT^(-2))/(L) =[M^1 L^ 0 T^(-2)]` `n_2 = n_1((M_1)/(M_2))^1 ((L_1)/(L_2))^0 ((T_1)/(T_2))^(-2) = 540 ((1g)/(1kg))^1 xx1((1s)/(60s))^(-2)` `n_2 = 540xx10^(-3)xx60xx60 = 1944` new units of surface tension |
|
| 242. |
The surface tension of water is 72 dyne//cm. Express is in SI units. |
|
Answer» Correct Answer - 0.072 N//m Surface tension, `sigma = (F)/(I) = [M^1 L^0T^(-2)]` Here, `n_1 = 72, M_1 = 1g, T_1 = 1s` `n_2 =?, M_2 =1kg, T_2 =1s` `n_2 = n_1((M_1)/(M_2))^1((T_1)/(T_2))^(-2) = 72 ((1g)/(1kg))^1((1s)/(1s))^(-2)` `=72xx10^(-3) =0.072N//,` |
|
| 243. |
The density of wood is `0.5g//cc` what is its value is SI? |
| Answer» Correct Answer - `500 kg//m^3` | |
| 244. |
A voltmeter having least count 0.1 V and an ammeter having least count 0.2 A are used to measure the potentail difference across the ends of a wire and current flowing through the wire respectively. If the reading of voltmeter is 4.4 V and reading of ammeter is 2.2 A, then find (i) the resistance of wire with maximum permissible error and (ii) maximum percentage error. |
|
Answer» Here, `V = 4.4 V, I = 2.2 A, Delta V = 0.1 V, DeltaI = 0.2 A` `R = (V)/(I) = (4.4)/(2.2) = 2 ohm` Maximum permissible error `(DeltaR)/(R ) = (DeltaV)/(V) +(DeltaI)/(I) = (0.1)/(4.4) +(0.2)/(2.2) = 0.023 + 0.091 = 0.114` `DeltaR= 0.114xxR = 0.114xx2 = 0.228` `DeltaR = 0.2 Omega`(rounding off to one decimal place) :. Resistance of wire with max. permissible error, `R = (2.0 +- 0.2)` ohm Max. percentage error `= (DeltaR)/(R )xx100 = 0.114xx100= 11.4%` |
|
| 245. |
A calorie is a unit of heat or energy and it equals about `4.2 J, where 1 J = 1 kg m^(2) s^(-2)`. Suppose we employ a system of units in which the unit of mass equals `alpha kg`, the unit of length equals is `beta m` , the unit of time is `gamma s`. Show tthat a calorie has a magnitude `4.2 alpha^(-1) beta^(-1) gamma^(2)` in terms of the new units. |
|
Answer» Here, 1 calorie = `4.2 J = 4.2 kg m^2 s^(-2)` As new unit of mass ` = alpha kg :. 1kg = (1)/(alpha)` new units of mass `= alpha^(-1)` new units of mass Similarly, `1 m = beta^(-1)` new unit of time Putting these value in (i), we obtain 1 calorie `= 4.2 (alpha^(-1) "new unit of mass ")(beta^(-1)" new unit of length")^2 (lambda^(-1) "new unit of times")^(-2)` `=4.2 alpha^(-1) beta^(-2) lambda^2` new unit of energy, which was to be proved. |
|
| 246. |
Find the dimensions of `axxb` in the relation ` p= a sqrtt - bx^2`, where x is distance, t is time and P is power. |
|
Answer» Correct Answer - `M^2L^2T^(-13//2)` `P = a sqrt - bx^2` `asqrtt = P = ML^2T^(-3)` `a =(P)/(sqrtt) =ML^2T^(-7//2)` Again, `bx^2 = P, b=(P)/(x^2) = (ML^2T^(-3))/(L^2) = MT^(-3)` `axxb = (ML^2T^(-7//2))xx(MT^(-3))` `=M^2L^2 T^(-13//2)` |
|
| 247. |
Find the dimensions of axxb in the relation ` P= (a -t^2)/(bsqrtx)`, wher x is distance t is time and P is power. |
|
Answer» Correct Answer - `M^(-1)L^(-5//2)T^7` `P = (a -t^2)/(bsqrtx)` Dimension of a = `[T^2]` Again, `(T^2)/(bsqrtx) =P, b =(T^2)/(Psqrtx) = (T^2)/([ML^2 T^(-3)]L^(1//2))` `b = M^(-1) L^(-5//2) T^5` `:. Axxb =T^2xxM^(-1)L^(-5//2) T^5= [M^(-1)L^(-5//2) T^7]` |
|
| 248. |
Write the dimensions of a and b in the relation ,` P = (b-x^(2))/(at)` , where P is power ,x is distance and t is timeA. `[M^0 lT^(-2)]`B. `[M^0L^2 T^2]`C. `[M^0 L^2 T^(-2)]`D. `[M^0 L^2 T^0]` |
|
Answer» Correct Answer - (d) As `x^2` is subtracted from b, therefore, dimensional formula of `b = x^2 = L^2 = [M^0 L^2 T^0]` |
|
| 249. |
Whether a given relation / formula is correct or not can be checked on the basis of the principle of homogeneity of dimensions. According to this principle, only that formula is correct, in which the dimensions of the various terms on one side of the relation are equal to the respective dimensions of these terms on the other side of the relation. With the help of the comprehension given above, choose the most appropriate alternative for each of the following questions : In the same equation, the dimensional formula of a isA. `[M^(-1) L^0 T^2]`B. `[ML^0 T^2]`C. `[ML^(-1) T^(-2)]`D. `[M^(-1) L^1 T^(-2)]` |
|
Answer» Correct Answer - (a) From the given relation, `a =(b-x^2)/(Pt) = (L^2)/([mL^2 T^(-3)])xxT = [M^(-1) L^0 T^2]` |
|
| 250. |
What are the dimensions of a and b in the relation F = at + b x, whrer F is force and x is distance ? |
|
Answer» `a = (F)/(t) = (MLT^(-2))/(T) = [M^1 L^1 T^(-3)],` `b = (F)/(x) =(MLT^(-2))/(L) = [M^1 L^0 T^(-2)]` |
|