100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	What is the maximum sum of the terms in the arithmetic progression 25, 24, 23, 22……? (a) 325 (b) 345 (c) 332.5 (d) 350
Answer» Correct option (a) 325 Explanation: The maximum sum would occur when we take the sum of all the positive terms of the series. The series 25, 24, 23,… 1, 0 has 26 terms. The sum of the series would be given by: n × average = 26 × 12.5 = 325.

1.

What is the maximum sum of the terms in the arithmetic progression 25, 24, 23, 22……? (a) 325 (b) 345 (c) 332.5 (d) 350

Answer»

Correct option (a) 325

Explanation:

The maximum sum would occur when we take the sum of all the positive terms of the series. The series 25, 24, 23,… 1, 0 has 26 terms. The sum of the series would be given by:

n × average = 26 × 12.5 = 325.

Discussion

2.	How many terms are identical in the two A.P.s 21, 23, 25,... up to 120 terms and 23, 26, 29,... up to 80 terms? (a) 39 (b) 40 (c) 41 (d) None of these.
Answer» Correct option (b) 40 Explanation: The first common term is 23, the next will be 29 (Notice that the second common term is exactly 6 away from the first common term. 6 is also the LCM of 2 and 3 which are the respective common differences of the two series.) Thus, the common terms will be given by the A.P 23, 29, 35 ....., last term. To find the answer you need to find the last term that will be common to the two series. The first series is 23, 25, 27 ... 259 While the second series is 23, 26, 29 ..... 260. Hence, the last common term is 257 Thus our answer becomes [(257 -23)/6]+1 =40.

2.

How many terms are identical in the two A.P.s 21, 23, 25,... up to 120 terms and 23, 26, 29,... up to 80 terms? (a) 39 (b) 40 (c) 41 (d) None of these.

Answer»

Correct option (b) 40

Explanation:

The first common term is 23, the next will be 29 (Notice that the second common term is exactly 6 away from the first common term. 6 is also the LCM of 2 and 3 which are the respective common differences of the two series.)

Thus, the common terms will be given by the A.P 23, 29, 35 ....., last term. To find the answer you need to find the last term that will be common to the two series.

The first series is 23, 25, 27 ... 259

While the second series is 23, 26, 29 ..... 260.

Hence, the last common term is 257

Thus our answer becomes [(257 -23)/6]+1 =40.

Discussion

3.	An electric field converges at the origin whose magnitude is given by the expression `E=100r N//C`, where r is the distance measured from the origin.A. Total charge contained in any spherical volume with its centre at origin negative.B. Total charge contained at any spherical volume, irrespective of the location of its centre, is negativeC. Total charge contained in a spherical volume of radius 3 cm with its centre at origin equals `3xx10^(-13) C`D. Total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude `3xx10^(-9) C`.
Answer» Correct Answer - A::B::C As the electric field converges at the origin so total charge contained in any spherical volume, irrespective of the location, is negative. By Gauss theorem `int vec(E).dvec(s)=q/in_(0)` We have `-E(4pir^(2))=q/in_(0)rArr q=-3xx10^(-13) C`

Discussion

4.	A radioactive sample has `6.0 xx 10^18` active nuclei at a certain instant. How many of these nuclei will still be in the same active state after two half-lives?A. `1.5 xx 10^(18)`B. `3 xx 10^(18)`C. `6 xx 10^(18)`D. None of these
Answer» Correct Answer - A In one half-life the number of active nuclei reduces to half to original number. Thus in two half lives the number is reduced to `((1)/(2))xx((1)/(2))` of the original number. The number of remaining active nuclei is therefore. `6.0 xx 10^(18) xx ((1)/(2))xx((1)/(2)) = 1.5 xx 10^(18)`.

Discussion

5.	A bullet of mass m and charge q is fired towards a solid uniformly charge sphere of radius R and total charge +q. If it strikes the surface of sphere with speed u, find the minimum value of u so that it can penetrate through the sphere. (Neglect all resistance force or friction acting on bullet except electrostatic forces) A. `q/sqrt(2piepsilon_(0)mR)`B. `q/sqrt(4piepsilon_(0)mR)`C. `q/sqrt(8piepsilon_(0)mR)`D. `(sqrt(3)q)/sqrt(4piepsilon_(0)mR)`
Answer» Correct Answer - B Energy at surface = Energy at centre `1/2 m u^(2)+(Kqxxq)/R=3/2 (Kq)/(R)xxq+0` `:. U=q/sqrt(4pi in_(0)mR)`

Discussion

6.	A unit positive point charge of mass m is projected with a velocity v inside the tummel as shown. The tunnel has been made inside a uniformly charged non conducting sphere. The minimum velocity with which the point charge should be projected such that it can it reach the opposite end of the tunnel, is equal to: A. `[(rhoR^(2))/(4mepsilon_(0))]^(1//2)`B. `[(rhoR^(2))/(24mepsilon_(0))]^(1//2)`C. `[(rhoR^(2))/(6mepsilon_(0))]^(1//2)`D. zero because the initial and the final points are at same potential
Answer» Correct Answer - A `(KQ)/(R)+1/2m u^(2)=(KQ)/(2R^(3))(3R^(2)-R^(2)/4)+0` `(1)/(4pi in_(0)R)xxrhoxx4/3piR^(3)+1/2 m u^(2)=(11xxrhoxx4/3piR^(3))/(8xx2R^(2))` `:.u=[(rhoR^(2))/(4m in_(0))]^(1//2)`

Discussion

7.	Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under `Kettle A: ` specific heat capacity `= 1680 Jkg^(-1)K^(-1)` Mass = 200 g Cost = Rs. 400 Kettle B: Specific heat capacity = `2450 Jkg^(-1) K^(-1)` Mass = 400 g Cost = Rs. 400 When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as (Energy used for liquid heating)/(Total energy supplied) They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)` Efficiency of kettle B is .A. `82.5%`B. `72.5%`C. `92.5%`D. `62.5%`
Answer» Correct Answer - D b. `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 4.8(theta-theta_1)) xx 100 = 250//3 = 83.34%` `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 6.4(theta-theta_1)) xx 100 = 62.5%` `H_(A) = 0.2 xx 1680 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx [0.8 +4] (theta - theta_1)` `H_(B) = 0.4 xx 2450 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx (2.4 + 4)(theta-theta_1)` . `H_(A)/H_(B) = 4.8/6.4 = 3/4` `H_(A) = V^2/R_(A) xx 6` `H_(B) = V^2/R_(B) xx 8` `H_(A)/H_(B) = R_(B)/R_(A) xx 3/4 but H_(A)/H_(B) = 3/4 ` Then `R_(B) = R_(A)`

Discussion

8.	Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under `Kettle A: ` specific heat capacity `= 1680 Jkg^(-1)K^(-1)` Mass = 200 g Cost = Rs. 400 Kettle B: Specific heat capacity = `2450 Jkg^(-1) K^(-1)` Mass = 400 g Cost = Rs. 400 When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as (Energy used for liquid heating)/(Total energy supplied) They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)` Ratio of efficiency consumed charges for one time boiling of tea in kettle A to that in kettle B isA. `3:5`B. `2:3`C. `3:4`D. `1:1`
Answer» Correct Answer - C c. `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 4.8(theta-theta_1)) xx 100 = 250//3 = 83.34%` `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 6.4(theta-theta_1)) xx 100 = 62.5%` `H_(A) = 0.2 xx 1680 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx [0.8 +4] (theta - theta_1)` `H_(B) = 0.4 xx 2450 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx (2.4 + 4)(theta-theta_1)` . `H_(A)/H_(B) = 4.8/6.4 = 3/4` `H_(A) = V^2/R_(A) xx 6` `H_(B) = V^2/R_(B) xx 8` `H_(A)/H_(B) = R_(B)/R_(A) xx 3/4 but H_(A)/H_(B) = 3/4 ` Then `R_(B) = R_(A)`

Discussion

9.	Ram ans Shyam purchased two electric tea kettles A and B of same size, same thickness, and same volume of 0.4L. They studied the specification of kettles as under `Kettle A: ` specific heat capacity `= 1680 Jkg^(-1)K^(-1)` Mass = 200 g Cost = Rs. 400 Kettle B: Specific heat capacity = `2450 Jkg^(-1) K^(-1)` Mass = 400 g Cost = Rs. 400 When kettle A is switched on with constant potential source, the tea begins to boil in 6 min. When kettle B is switched on with the same source separately, then tea begins to boil in 8 min. The efficiengy of kettle is defined as (Energy used for liquid heating)/(Total energy supplied) They made discussion on specification and efficiency of kettles and subsequently prepared a list of questions to draw the conclusions. Some of them are as under (assume specific heat of tea liquid as `4200J kg^(-1)K^(-1)` and density `1000 kgm^(-3)` If resistances of coil of kettles A and B are `R_(A) and R_(B)`, respectively, then we can sayA. `R_AgtR`B. `R_A = R_B`C. `R_AltR_B`D. cannot be ascertained by above data.
Answer» Correct Answer - B b. `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 4.8(theta-theta_1)) xx 100 = 250//3 = 83.34%` `eta = (0.4 xx 4200 (theta - theta_1))/(420 xx 6.4(theta-theta_1)) xx 100 = 62.5%` `H_(A) = 0.2 xx 1680 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx [0.8 +4] (theta - theta_1)` `H_(B) = 0.4 xx 2450 (theta-theta_1) + 0.4 xx 4200 (theta-theta_1)` `= 420 xx (2.4 + 4)(theta-theta_1)` . `H_(A)/H_(B) = 4.8/6.4 = 3/4` `H_(A) = V^2/R_(A) xx 6` `H_(B) = V^2/R_(B) xx 8` `H_(A)/H_(B) = R_(B)/R_(A) xx 3/4 but H_(A)/H_(B) = 3/4 ` Then `R_(B) = R_(A)`

Discussion

10.	Statement I: A wire of uniform cross-section and uniform resistivity is connected across an ideal cell. Now the length of the wire is doubled keeping volume of the wire constant. The drift velocity of electrons after stretching the wire becomes one-fouth of what it was before stretching the wire. Statement II: If a wire (of uniform resistivity and uniform cross section) of length `l_0` is stretched to length `nl_0`, then its resistance becomes `n^2` times of what it was before stretching the wire (the volume of wire is kept constant in stretching process). Further at constant potential difference, current is inversely proportional to resistance. Finally, drift velocity of free electron is directly proportional to current and inversely proportional to cross-sectional area of current carrying wire.A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.
Answer» Correct Answer - D d. As the length of the wire is doubled, the cross-sectional area of the wire becomes half. Therefore, resistance of the wire becomes four times and the current beocmes one- fourth of the initial value. Also, `V_(d) = I/(neA)` Since current becomes one-fourth and cross-sectional area of the wire becomes half, from the above equation the drift velocity of electron becomes half. Hence, Statement I is false.

Discussion

11.	In the given circuit , `R_1 != R_2` and the reading of the voltmeter is the same, irrespective of whether the switch S is open or closed. Then, which of the following is correct? A. `I_(R_2) = I_V`B. `I_(R_1) = I_(R_2)`C. `I_(R_3) = I_V`D. none of these
Answer» Correct Answer - A a. This is a case of balanced Wheatstone bridge.

Discussion

12.	In a uniformly charged dielectric sphere, a very thin tunnel has been made along the diameter as shown in figure. A charge partivel `-q` having mass m is released from rest at one end of the tunnel. For the situation described, mark the correct statement(s), (negalect gravity). A. Charge particle will perform SHM about center of the sphere as mean position.B. Time period of the particle is `2pi sqrt(2piepsilon_0mR^3//qR)`C. particle will perform oscillation but not SHM.D. Speed of the particle while crossing the mean position is `sqrt (qQ//4piepsilon_0mR).`
Answer» Correct Answer - A::B::D a.,b.,d. At a center, the force experienced by the charge particle is zero, so it is a position of equilibrium. As we displace the charge so it is a position of equilibrium. As we displace the chare from the equilibrium position, electric force starts acting on it toward equilibrium position and hence equilibrium is a stable one. At a distance `x` from the center of sphere (equilibrium position), force experienced by the charge particle is `F=(qQx)/(4piepsilon_(0)R^(3))` [for `xltR`] As `F=x`, it performs `SHM` about the center. Time period can be calculated by using `F=mw^(2)x`.

Discussion

13.	In the given circuit , the potential difference across the capacitor is 12 V. Each resistance is of `3 Omega`. The cell is ideal. The emf of the cell is A. 15 VB. 9 VC. 12 VD. 24 V
Answer» Correct Answer - A a. In the steady state, no current flows through the capacitor.

Discussion

14.	A battery of internal resistance `4Omega` is connected to the network of the resistance as shown in figure . If the maximum power can be delivered to the network, the magnitude of resistance in `Omega`should be A. `19//21 Omega`B. `84//19 Omega`C. `12 Omega`D. `7 Omega`
Answer» Correct Answer - A a. Recall the condition for maximum power flow through a circuit.

Discussion

15.	100 ml `Fe(OH)_(3)` is coagulated by 10 ml 1N `Na_(2)SO_(4)`. Then what will be coagulation value of `Na_(2)SO_(4)`?A. 5B. 10C. 50D. 100
Answer» Correct Answer - C 100 ml `Fe(OH)_3 ` is coagulated by =`10xx1/2`=5 milli mol 1 mol `Fe(OH)_3` is coagulated by = `5/100` 1000 ml `Fe(OH)_3` is coagulated by `=(1000xx5)/100`=50

Discussion

16.	Which of the following options are CORRECT (ammeters are ideal). A. Reading of `A_(1) is 1.16 A`B. Reading of voltameter v is `77.14 V`C. Reading of ammeter `A_(2) is 0.77 A`D. Reading of voltmeter is `20 V`
Answer» Correct Answer - A&B&C

Discussion

17.	What is the total life of a radioactive element ?
Answer» Correct Answer - Infinity.

Discussion

18.	What is the source of radioactive `CO_(2)` in the atmosphere?
Answer» `._(7)N^(14)` is hit by cosmic ray neutrons to produce `._(6)C^(14)` which is radioactive and is oxidized to produce radioactive `XO_(2)`.

Discussion

19.	What are transactinides ? To which block do they belong ?
Answer» Elements with `Z gt103` are called transactinides. They belong to `d-`block.

Discussion

20.	Why is synthesis of transactinides difficult ?
Answer» The synthesis of transactinides is difficult because it is extermely expensive to build accelerators which can accelrate the medium weight nuclei required for bombardment.

Discussion

21.	Name the fundamental particle which exists in the nucleus with protons and neutrons.
Answer» Correct Answer - Measons.

Discussion

22.	30 gm of urea (M=60 gm /mol) is dissolved in 846 gm of water. Calculate the vapour pressure of water for this solution if the vapour pressure of pure water at 298 K 23.8 mm Hg
Answer» Given, weight of urea (W₂) = 30g Weight of water (W₁) = 846g Vapour pressure of water P₁^° = 23.8 mm Hg n_B = \(\frac{30}{60}\) = 0.5, n_A = \(\frac{846}{18}\) = 47 Mole fraction of water (x_A) = \(\frac{n_A}{n_A\ +\ n_B}\) =\(\frac{47}{47\ +\ 0.5}\) = \(\frac{47}{47.5}\) = 0.99 P_A= P_A ^° × x_A= 23.8 × 0.99 = 23.5 mm Hg.

22.

30 gm of urea (M=60 gm /mol) is dissolved in 846 gm of water. Calculate the vapour pressure of water for this solution if the vapour pressure of pure water at 298 K 23.8 mm Hg

Answer»

Given, weight of urea (W₂) = 30g

Weight of water (W₁) = 846g

Vapour pressure of water P₁^° = 23.8 mm Hg

n_B = \(\frac{30}{60}\) = 0.5, n_A = \(\frac{846}{18}\) = 47

Mole fraction of water (x_A) = \(\frac{n_A}{n_A\ +\ n_B}\)

=\(\frac{47}{47\ +\ 0.5}\)

= \(\frac{47}{47.5}\) = 0.99

P_A= P_A ^° × x_A= 23.8 × 0.99 = 23.5 mm Hg.

Discussion

23.	Write the IUPAC name and hybridization of the following complexes: (a) [Ni (CO)4] (b) [CoF6] 3- [ Atomic number of Co= 28, Ni = 27]
Answer» (a) Tetracarbonylnickel (0), hybridisation- sp³ (b) hexafluoridocobaltate (III), Hybridisation - sp³d²

23.

Write the IUPAC name and hybridization of the following complexes: (a) [Ni (CO)4] (b) [CoF6] 3- [ Atomic number of Co= 28, Ni = 27]

Answer»

(a) Tetracarbonylnickel (0), hybridisation- sp³

(b) hexafluoridocobaltate (III), Hybridisation - sp³d²

Discussion

24.	Which radioactive metal is used as an ingredient of atomic explosive weapons ?
Answer» Correct Answer - Uranium.

Discussion

25.	`C-14` is formed in the upper atmosphere. Why it does not occur at the ground level ?
Answer» Cosmic ray neutrons are present only in upper atmosphere and not at the ground leve.

Discussion

26.	Which amino acid can protonate and deprotonate at neutral pH?
Answer» Click on the structures below to switch between their protonated and deprotonated forms. For these amino acids, the protonated forms predominate at physiological pH (about 7). Two amino acids have acidic side chains at neutral pH. These are aspartic acid or aspartate (Asp) and glutamic acid or glutamate (Glu). ASPARTATE and GLUTAMATE

Discussion

27.	The method used to remove temporary hardness of water is :(1) Calgon's method(2) Clark's method(3) Ion-exchange method(4) Synthetic resins method
Answer» Correct option (2) Clark's method Explanation: Clark's method is used to remove temporary hardness of water, in which bicarbonates of calcium and magnesium are reacted with slaked lime Ca(OH)₂ Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ ↓+ 2H₂O Mg(HCO₃)₂ + 2Ca(OH)₂ → 2CaCO₃ ↓ + Mg(OH)₂ ↓ + 2H₂O

27.

The method used to remove temporary hardness of water is :(1) Calgon's method(2) Clark's method(3) Ion-exchange method(4) Synthetic resins method

Answer»

Correct option (2) Clark's method

Explanation:

Clark's method is used to remove temporary hardness of water, in which bicarbonates of calcium and magnesium are reacted with slaked lime Ca(OH)₂

Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ ↓+ 2H₂O

Mg(HCO₃)₂ + 2Ca(OH)₂ → 2CaCO₃ ↓ + Mg(OH)₂ ↓ + 2H₂O

Discussion

28.	Which one is malachite from the following ? (1) CuFeS2 (2) Cu(OH)2 (3) Fe3O4 (4) CuCO3.Cu(OH)2
Answer» Correct option (4) CuCO₃.Cu(OH)₂ _Explanation: Malachite : CuCO₃.Cu(OH)₂ (Green colour)

28.

Which one is malachite from the following ? (1) CuFeS2 (2) Cu(OH)2 (3) Fe3O4 (4) CuCO3.Cu(OH)2

Answer»

Correct option (4) CuCO₃.Cu(OH)₂

_Explanation:

Malachite : CuCO₃.Cu(OH)₂ (Green colour)

Discussion

29.	A 5-kg concrete block is lowered with a downward acceleration of 2.8m/s2 by means of a rope. The force of the block on the rope is
Answer» Consider all the forces along the x- axis: Fx = ma - mg Fx = m(a-g) Fx = 5 (2.8-9.8) Fx = 35 N down

29.

A 5-kg concrete block is lowered with a downward acceleration of 2.8m/s2 by means of a rope. The force of the block on the rope is

Answer»

Consider all the forces along the x- axis:

Fx = ma - mg

Fx = m*(a-g)

Fx = 5 *(2.8-9.8)

Fx = 35 N down

Discussion

30.	Compound that is of Prussian Blue Colour?(a) Na4[Fe(CN)6S](b) Na4[Fe(CN)6NCS](c) Na4[Fe(CN)5NOS](d) Na2[Fe(CN)5NOS]
Answer» (c) Na₄[Fe(CN)₅NOS] Na₂S+ Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS] Sodium sulphide reacts with sodium nitroprusside to form a violet colour compound, which confirms the presence of sulfur.

30.

Compound that is of Prussian Blue Colour?(a) Na4[Fe(CN)6S](b) Na4[Fe(CN)6NCS](c) Na4[Fe(CN)5NOS](d) Na2[Fe(CN)5NOS]

Answer»

(c) Na₄[Fe(CN)₅NOS]

Na₂S+ Na₂[Fe(CN)₅NO] → Na₄[Fe(CN)₅NOS]

Sodium sulphide reacts with sodium nitroprusside to form a violet colour compound, which confirms the presence of sulfur.

Discussion

31.	In an isolted, charged, parllel-plate air capacitor, the charge per unit area on each plate has a magnitude of `sigma`. A dielectric slab having the dielectric constant K is now introduced between the plates. The induced charge per unit area on the surface of the dielectric will have magnitude.A. `(sigma)/(K)`B. `sigma(K-1)`C. `sigma(1-(1)/(K))`D. `(sigma)/(K+1)`
Answer» Correct Answer - C

Discussion

32.	In solar radiation, the intensity of radiation is maximum around the wavelength `lambda_(m)`. If R is the radius of the sun and c is the velocity of light, the mass lost by the sum per unit time is proportional toA. `(R^(2))/(lambda^(4)c^(2))`B. `(R^(2))/(lambda^(2)c^(2))`C. `(R^(3))/(lambda^(4)c^(3))`D. `(R^(3))/(lambda^(4)c^(2))`
Answer» Correct Answer - A

Discussion

33.	A uniform rod of mass m, hinged at its upper end, is released from rest from a horizontal position. When it passes through the vertical position, the force on the hinge isA. `(3)/(2)mg`B. `2mg`C. `(5)/(2)mg`D. 3mg
Answer» Correct Answer - C

Discussion

34.	A wave is travelling along X-axis. The disturbance at x=0 and t=0 is `A//2` and is increasing. Where A is amplitute of the wave. If `y=A sin(kx-omegat+emptyset)`, deetemine the initial phase `empyset`.A. `(pi)/(6)`B. `(5 pi)/(6)`C. `(pi)/(3)`D. `(11 pi)/(6)`
Answer» Correct Answer - B At `t = 0, x = 0` `(A)/(2) = A sin phi rArr phi = (pi)/(6), (5pi)/(6)` since, `(dy)/(dt)` is positive for `phi = (5pi)/(6)`

Discussion

35.	What are the colors (of the electromagnetic spectrum) absorbed by plants? What would happen to photosynthesis if the green light waves that reach a vegetable were blocked?
Answer» Chlorophyll absorbs all other colours of the electromagnetic spectrum but it practically does not absorb the green. The green colour is reflected and such reflection provides the characteristic colour of plants. If the green light that reaches a plant is blocked and exposure of the plant to other colours is maintained there would be no harm to the photosynthesis process. Apparent paradox: the green light is not important for photosynthesis. There is a difference between the optimum colour frequency for the two main types of chlorophyll, the chlorophyll A and chlorophyll B. Chlorophyll A has an absorption peak at approximately 420 nm wavelength (anil) and chlorophyll B has its major absorption in 450 nm wavelength (blue).

35.

What are the colors (of the electromagnetic spectrum) absorbed by plants? What would happen to photosynthesis if the green light waves that reach a vegetable were blocked?

Answer»

Chlorophyll absorbs all other colours of the electromagnetic spectrum but it practically does not absorb the green. The green colour is reflected and such reflection provides the characteristic colour of plants. If the green light that reaches a plant is blocked and exposure of the plant to other colours is maintained there would be no harm to the photosynthesis process. Apparent paradox: the green light is not important for photosynthesis.

There is a difference between the optimum colour frequency for the two main types of chlorophyll, the chlorophyll A and chlorophyll B. Chlorophyll A has an absorption peak at approximately 420 nm wavelength (anil) and chlorophyll B has its major absorption in 450 nm wavelength (blue).

Discussion

36.	A progressive wave travels in a medium `M_(1)` and enters into another mediun `M_(2)` in which its speed decreases to `60%`. Then the ratio of the amplitude of the transmitted and the incident waves isA. `3//5`B. `3//4`C. `3//10`D. `1`
Answer» Correct Answer - B `v_(2) = (60)/(100)v_(1) = (3)/(5)v_(1)` `(A_(t))/(A_(i)) = (2v_(2))/(v_(1) + v_(2)) = (2(v_(2))//(v_(1)))/(1 + ((v_(2))/(v_(1)))) = (2(3//5))/(1 + (3//5)) = (3)/(4)`

Discussion

37.	An object is subjected to an acceleration `a = 4 + 3v`. It is given that the displacement `S = 0` when `v = 0`. The value of displacement when `v = 3 m//s` isA. `1 + (4)/(3) ln((13)/(4))`B. `1 - (4)/(3)ln ((13)/(8))`C. `1 - (4)/(9) ln ((13)/(4))`D. `1 + (4)/(9) ln ((13)/(8))`
Answer» Correct Answer - C `a = v (dv)/(dS) = 4 + 3v` `underset(0)overset(v)int (vdv)/(4 + 3v) = underset(0)overset(S)int dS` `S = 1 - (4)/(9)l ((13)/(4))`

Discussion

38.	An unknown resistance r si determined in terms of a standared resistance `R = 100 Omega` by using potentiometer. The potential difference across r is balanced againt `45 cm` length of the wire and that `(r + R)` is obtained at `70 cm` of the wire. Find the value of the unknown resistance.A. `200 Omega`B. `280 Omega`C. `180 Omega`D. `100 Omega`
Answer» Correct Answer - C `l_(1) = 45 cm, l_(2) = 70 cm` `R = 100 Omega` `r = R ((l_(1))/(l_(2) - l_(1))) = 100 ((45)/(75 - 45)) = 180 Omega`

Discussion

39.	The increasing order of the deviation of the colours observed in a spectrum through a prism isa) Red, orange, yellow, green, blue, Indigo and violet.b) Violet, Indigo, Blue, green, yellow, orange, red.c) Red orange, green, blue, yellow, Indigo, violet.d) Violet, Indigo, green, yellow, blue, orange, red
Answer» a) Red, orange, yellow, green, blue, Indigo and violet

Discussion

40.	A force of `20 N` is applied on upper block as shown in the figure. Te total work done by frictional during the time interval in which the upper block has a displacement of `15 m` with respect to the ground is (take `g = 10 m//s^(2)` ) A. `50 J`B. `-50 J`C. `75 J`D. `- 75 J`
Answer» Correct Answer - B Block and plank both slide so, `a_("Block") = (20 -5)/(2) = (15)/(2)` `a_("plank") = (5)/(2)` `S = ut + (1)/(2)at^(2)` `15 = 0 + (1)/(2) (15)/(2)t^(2)` `rArr t = 2 sec` Desplacement of plane `= 0 + (1)/(2) (5)/(2)2^(2)` So, displacement of block with respect to plank `= 15 - 5 = 10m` `W_(f) = - 10 xx 20 xx 1//4 = -50J`

Discussion

41.	Which of the following chemical equation represents double displacement reaction.a) 2Mg + O2 → 2MgOb) CaCO3 \(\overset{heat}{\longrightarrow}\) CaO + CO2c) CuSO4 + Zn → ZnSO4 + Cud) AgNO3 + NaCl → AgCl + NaNO3
Answer» d) AgNO₃ + NaCl → AgCl + NaNO₃

Discussion

42.	Draw the output waveform across the resistor (Fig.) A. B. C. D.
Answer» Correct Answer - A The `p - n` junction diode acts as a half wave rectifier, so the output waveform across the resistor is positvie half part.

Discussion

43.	What is the path followed by the energy absorbed by plants to be used in photosynthesis?
Answer» The energy source of photosynthesis is the sun, the unique and central star of our planetary system. In photosynthesis, solar energy is transformed into chemical energy, the energy of the chemical bonds of the produced glucose molecules (and of the released molecular oxygen). The energy of glucose is then stored as starch (a glucose polymer) or it is used in the cellular respiration process and transferred to ATP molecules. ATP is consumed in metabolic processes that spend energy (for example, inactive transport across membranes).

Discussion

44.

Write any three differences between diamagnetic and paramagnetic materials.

Answer»

Diamagnetic	Paramagnetic
1. Weakly magnetised in a direction opposite to the applied magnetic field	1. Weakly magnetised along the direction of the applied magnetic field.
2. Move from stronger to weaker part of the external magnetic field	2. Move from weaker to the stronger part of the external magnetic field
3. Magnetic susceptibility is Low and negative	3. Magnetic susceptibility is low and positive.
4. Example: Bismuth, copper, lead, silicon	4. Example: Aluminium, sodium, calcium, oxygen

Discussion

45.	The vernier constant of a vernier callipers is `0.001 cm`. If `49` main scale divisions coincide with `50` vernier scale devisions, then the value of `1` main scale divisions is .A. `0.1mm`B. `0.5mm`C. `0.4mm`D. `1mm`
Answer» Correct Answer - B `50VSD=49MSD` `MSD=(49)/(50)MSD` `LC=1MSD-1VSD` `=1MSD-(49)/(50)MSD` or `LC=(1)/(50)MSD` `1 MSD=50(LC)` `=50(0.001)cm` `=0.05cm` `=0.5mm`.

Discussion

46.	In a vernier callipers, `N` divisions of the main scale coincide with `N + m` divisions of the vernier scale. what is the value of `m` for which the instrument has minimum least count.A. `1`B. `N`C. `(N)/(10)`D. `(N)/(2)`
Answer» Correct Answer - A `(N+m)VSD=(N)MSD` `:. 1VSD=((N)/(N+m))MSD` `LC=1MSD-1VSD` `=1MSD-((N)/(N+m))MSD` `=((1)/(1+N//m))MSD` For minimum value of least count `m` should be minimum or `1`.

Discussion

47.	`1 cm` on the main scale of a vernier callipers is divided into `10 equal` parts. If `10` divisions of vernier coincide with `8` small divisions of main scale, then the least count of the calliper is.A. `0.01 cm`B. `0.02 cm`C. `0.05 cm`D. `0.005 cm`
Answer» Correct Answer - B `10VSD=8MSD` `1VSD=0.8MSD` Now `LC=1MSD-1VSD` `=1MSD-0.8MSD` `0.2MSD=0.2xx(1)/(10)cm` `=0.02cm`.

Discussion

48.	The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (-0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be_____ x 10-2 cm.
Answer» Correct answer is 180 Measured diameter = MSR + VSR × VC = 1.7 + 0.01 × 5 = 1.75 Corrected = Measured – Error = 1.75 – (–0.05) = 1.80 cm = 180 x 10^–2 cm 180

48.

The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (-0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be_____ x 10-2 cm.

Answer»

Correct answer is 180

Measured diameter = MSR + VSR × VC

= 1.7 + 0.01 × 5 = 1.75

Corrected = Measured – Error

= 1.75 – (–0.05)

= 1.80 cm

= 180 x 10^–2 cm

180

Discussion

49.	द्रव्यमान m का एक आवेशित कण (आवेश q ) l लम्बाई के किसी हल्के अवितानय (inextensible ) धागे से बंधा है तथा दूसरा सिरा बिन्दु O से बंधा है। सम्पूर्ण व्यवस्था किसी चिकने क्षैतिज तल पर स्थित है (चित्र 2.2 )। प्रारम्भ में कोण की स्थिति A पर है और उसी समय एकसमान विधुत-क्षेत्र `vecE,OB` दिशा के समान्तर स्थापित किया जाता है। A. बिन्दु B पर कण की चाल `=sqrt((qEl)/(m))`B. बिन्दु B पर कण की चाल `=sqrt((2qEl)/(m))`C. बिन्दु B पर धागे में तनाव शून्य हैD. बिन्दु B पर धागे में तनाव 2qE है
Answer» Correct Answer - A::D विधुत-क्षेत्र द्वारा किया गया कार्य, `W=(qE)(AB)(cos60^(@)),` कार्य-ऊर्जा प्रमेय से `1/2mv^(2)1/2qEl,` फिर बिन्दु B पर आवश्यक अभिकेन्द्र बल `(mv^(2))/(l) =`तनाव (T)-qE.

Discussion

50.	एक-दूसरे के समांतर व्यस्थित दो विशाल सुचालक प्लेट X तथा Y को क्रमशः `Q_(1)` और `Q_(2)` आवेश दिया गया है। चित्र 2.6 में A, B, C, D द्वारा प्रदर्शित है। A. A और D prastho पर आवेश के परिणाम भिन्न होंगे।B. A और D दोनों पर समान आवेश `1/2(Q_(1)+Q_(2))` होगा।C. B और C पर आवेश परिणाम एव कृत प्रकृति दोनों में भिन्न होंगे।D. B पर आवेश `1/2(Q_(1)-Q_(2))` तथा C आवेश `1/2(Q_(2)-Q_(1))` होगा।
Answer» Correct Answer - B::D देखे खंड 1 में धारा 2.7 : गॉस के प्रमेय के अनुप्रयोग का भाग 6

Discussion

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