100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 126 InterviewSolution

6251.	___________ algorithm keeps track of k states rather than just one.(a) Hill-Climbing search(b) Local Beam search(c) Stochastic hill-climbing search(d) Random restart hill-climbing search
Answer» The correct answer is (b) Local Beam search To explain: Refer to the definition of Local Beam Search algorithm.

Discussion

6252.	(+)-2-Butanol has [alpha]20D=+13.90. A sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of -3.50 under similar conditions. The percentages of the(+) and (-) enantiomers present in the mixture are, respectively.
Answer» Given (+)-2-Butanol has [alpha]²⁰D=+13.9⁰ ^{So its enantiomer}(-)-2-Butanol has [alpha]²⁰D=-13.9⁰ ^Nowa sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of -3.5⁰ under similar conditions. So itis due to the presence excess (-)-2-Butanol. Hence the excess percentage of (-)-2-Butanol in the mixture will be = (3.5x100)/13.9 ≈ 25.18%. The remainder of the sample is a racemic mixture of the enantiomers and in the racemic mixture each is present in (100-25.18)/2%=37.41% Hence we get the percentages of the(+) and (-) enantiomers present in the mixture are, respectively.as (+)-2-Butanol=37.41% and (-)-2-Butanol=37.41%+25.18%=62.59%

6252.

(+)-2-Butanol has [alpha]20D=+13.90. A sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of -3.50 under similar conditions. The percentages of the(+) and (-) enantiomers present in the mixture are, respectively.

Answer»

Given (+)-2-Butanol has [alpha]²⁰D=+13.9⁰

^{So its enantiomer}(-)-2-Butanol has [alpha]²⁰D=-13.9⁰

^Nowa sample of 2-butanol containing both the enantiomers was found to have a specific rotation value of -3.5⁰ under similar conditions. So itis due to the presence excess (-)-2-Butanol. Hence the excess percentage of (-)-2-Butanol in the mixture will be = (3.5x100)/13.9 ≈ 25.18%. The remainder of the sample is a racemic mixture of the enantiomers and in the racemic mixture each is present in (100-25.18)/2%=37.41%

Hence we get the percentages of the(+) and (-) enantiomers present in the mixture are, respectively.as

(+)-2-Butanol=37.41% and (-)-2-Butanol=37.41%+25.18%=62.59%

Discussion

6253.	Hill-Climbing approach stuck for which of the following reasons?(a) Local maxima(b) Ridges(c) Plateaux(d) All of the mentioned
Answer» Correct option is (d) All of the mentioned For explanation: Local maxima: a local maximum is a peak that is higher than each of its neighboring states, but lower than the global maximum. Ridges: Ridges result in a sequence of local maxima that is very difficult for greedy algorithms to navigate. Plateaux: a plateau is an area of the state space landscape where the evaluation function is flat.

Discussion

6254.	Hill climbing sometimes called ____________ because it grabs a good neighbor state without thinking ahead about where to go next.(a) Needy local search(b) Heuristic local search(c) Greedy local search(d) Optimal local search
Answer» The correct option is (c) Greedy local search To explain: None.

Discussion

6255.	Which of the following rae state functions? (a) Height of a hill (b) Distance travelled in climbing the hill (c ) Energy change in climbing the hill
Answer» Energy consumed in climbing the hill.

Discussion

6256.	The number of moles of `KMnO_(4)` that will be needed to react with `1 mol` of sulphite ion in acidic solution isA. `4//5`B. `2//5`C. 1D. `3//5`
Answer» Correct Answer - B `2KMnO_(4)+3H_(2)SO_(4)toK_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[O]` `[MnO_(4)^(-)+8H^(+)+5e^(-)toMn^(2+)+4H_(2)O]xx2` `[SO_(3)^(2-)+H_(2)OtoSO_(4)^(2-)+2H^(+)+2e^(-)]xx5` `2MnO_(4)^(-)+6H^(+)+5SO_(3)^(2-)to2Mn^(2+)+5 SO_(4)^(2-)+3H_(2)O` 5 moles of sulphite ions react with =2 moles of `MnO_(4)^(-)` So , 1 mole of sulphite ions react with `=(2)/(5)" moles of "MnO_(4)^(-)`.

Discussion

6257.	An element X has the following isotopic composition: `.^(200)X:90% .^(199)X:8.0% .^(202)X:2.0%` The weight average atomic mass of the naturally occurring element X is closest toA. 201uB. 202 uC. 199 uD. 200 u
Answer» Correct Answer - D Weight of `^.(200)X=0.90xx200=180.00u` Weight of `.^(199)X=0.08xx199=15.92u` Weight of `.^(202)X=0.02xx202=4.04u` Total weight `=199.96~~200u`

Discussion

6258.	Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3.The probability that x1 , x2 , x3 are in an arithmetic progression, is(A) 9/105(B) 10/105(C) 11/105(D) 7/105
Answer» (C) 11/105 Here 2x₂ = x₁ + x₃ ⇒ x₁ + x₃ = even Hence number of favourable ways = ²C₁ . ⁴C₂ + ¹C₁ . ³C₁ = 11.

6258.

Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3.The probability that x1 , x2 , x3 are in an arithmetic progression, is(A) 9/105(B) 10/105(C) 11/105(D) 7/105

Answer»

(C) 11/105

Here 2x₂ = x₁ + x₃

⇒ x₁ + x₃ = even

Hence number of favourable ways = ²C₁ . ⁴C₂ + ¹C₁ . ³C₁ = 11.

Discussion

6259.	Let bi >1 for i = 1, 2, ..., 101. Suppose loge b1, loge b2, , loge b101 are in Arithmetic Progression (A. P.) with the common difference loge 2. Suppose a1, a2, a101 are in A.P. such that a1 b1 and as1 b51. If = t = b1 + b2 +...+ b51 and s = a1 + a2 + ... + a51· then (A) s > t and a101 > b101 (B) s > t and a101 < b101 (C) s < t and a101 > b101 (D) s < t and a101 < b101
Answer» (B) s > t and a₁₀₁ < b₁₀₁ a₂, a₃, ....., a50 are Arithmetic Means and b₂, b₃, ....., b50 are Geometric Means between a₁(= b₁) and a₅₁(= b₅₁) Hence b₂ < a₂, b₃ < a₃ ..... ⇒ t < S Also a₁, a₅₁, a101 is an Arithmetic Progression and b₁, b₅₁, b₁₀₁ is a Geometric Progression Since a₁ = b₁ and a₅₁ = b₅₁ ⇒ b₁₀₁ > a₁₀₁

6259.

Let bi >1 for i = 1, 2, ..., 101. Suppose loge b1, loge b2, , loge b101 are in Arithmetic Progression (A. P.) with the common difference loge 2. Suppose a1, a2, a101 are in A.P. such that a1 b1 and as1 b51. If = t = b1 + b2 +...+ b51 and s = a1 + a2 + ... + a51· then (A) s > t and a101 > b101 (B) s > t and a101 < b101 (C) s < t and a101 > b101 (D) s < t and a101 < b101

Answer»

(B) s > t and a₁₀₁ < b₁₀₁

a₂, a₃, ....., a50 are Arithmetic Means and b₂, b₃, ....., b50 are Geometric Means between a₁(= b₁) and a₅₁(= b₅₁)

Hence b₂ < a₂, b₃ < a₃ .....

⇒ t < S

Also a₁, a₅₁, a101 is an Arithmetic Progression and b₁, b₅₁, b₁₀₁ is a Geometric Progression

Since a₁ = b₁ and a₅₁ = b₅₁

⇒ b₁₀₁ > a₁₀₁

Discussion

6260.	The increasing order of atomic radii of the following Group 13 elements is (A) Al < Ga < In < Tl (B) Ga < Al < In < Tl (C) Al < In < Ga < Tl (D) Al < Ga < T1 < In
Answer» (B) Ga < Al < In < Tl Increasing order of atomic radius of group 13 elements Ga < Al < In < Tl. Due to poor shielding of d-electrons in Ga, its radius decreases below Al.

6260.

The increasing order of atomic radii of the following Group 13 elements is (A) Al < Ga < In < Tl (B) Ga < Al < In < Tl (C) Al < In < Ga < Tl (D) Al < Ga < T1 < In

Answer»

(B) Ga < Al < In < Tl

Increasing order of atomic radius of group 13 elements Ga < Al < In < Tl. Due to poor shielding of d-electrons in Ga, its radius decreases below Al.

Discussion

6261.	If the sum of the zeros of the polynomial x2 - (k + 3)x + (5k - 3) is equal to one fourth of the product of its zeros find the value of k.
Answer» Given polynomial is x² - (k + 3)x + (5k - 3) Given that sum of zeros = \(\frac14\) x product of zeros i.e., \(\frac{-(-(k +3))}{1} = \frac14 \times \frac{5k -3 }{1}\) ⇒ 4k + 12 = 5k - 3 ⇒ k = 12 + 3 = 15

6261.

If the sum of the zeros of the polynomial x2 - (k + 3)x + (5k - 3) is equal to one fourth of the product of its zeros find the value of k.

Answer»

Given polynomial is x² - (k + 3)x + (5k - 3)

Given that sum of zeros = \(\frac14\) x product of zeros

i.e., \(\frac{-(-(k +3))}{1} = \frac14 \times \frac{5k -3 }{1}\)

⇒ 4k + 12 = 5k - 3

⇒ k = 12 + 3 = 15

Discussion

6262.	Write the mechanism of formation of photochemical smog.
Answer» At high temp, the petrol and diesel engines, N₂ & O₂ combine to form NO which is emitted into atmosphere. NO is then oxidised in air to form NO₂ which absorbs sunlight and form NO and free O atom. NO₂(g) + sunlight → NO(g) + O(g) The O atoms being reactive and combine with O₂ to form O₃ . O₂(g) + O(g) → O₃(g) The O₃react with NO formed by the photochemical decomposition of NO₂ . NO(g) + O₃(g) → NO₂(g) + O₂(g) NO₂ and O₃ are good oxidising agents and they react with unburnt hydrocarbons in the polluted air to form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

6262.

Write the mechanism of formation of photochemical smog.

Answer»

At high temp, the petrol and diesel engines, N₂ & O₂ combine to form NO which is emitted into atmosphere. NO is then oxidised in air to form NO₂ which absorbs sunlight and form NO and free O atom.

NO₂(g) + sunlight → NO(g) + O(g)

The O atoms being reactive and combine with O₂ to form O₃ .

O₂(g) + O(g) → O₃(g)

The O₃react with NO formed by the photochemical decomposition of NO₂ .

NO(g) + O₃(g) → NO₂(g) + O₂(g)

NO₂ and O₃ are good oxidising agents and they react with unburnt hydrocarbons in the polluted air to form substances such as acrolein and formaldehyde. These are the main substances of photochemical smog.

Discussion

6263.	For your agricultural field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer» The compost producing pit should be set up at a suitable place or in a tin to protect ourselves from bad odour and flies. It should be kept covered so that flies cannot make entry into it and the bad odour is minimized. The recyclable material like plastics, glass, newspapers, etc., should be sold to the vendor who further sells it to the dealer. The dealer further supplies it to the industry involved in recycling process.

Discussion

6264.	If A = \(\begin{bmatrix}x&y\\[0.3em]z&w\end{bmatrix},\) B = \(\begin{bmatrix}x&-y\\[0.3em]-z&w\end{bmatrix}\) and C = \(\begin{bmatrix}-2x&0\\[0.3em]0&-2w\end{bmatrix},\) then A + B + C is a(a) identity matrix (b) null matrix (c) row matrix (d) column matrix
Answer» (b) A + B + C = \(\begin{bmatrix}x&y\\[0.3em]z&w\end{bmatrix}\)+ \(\begin{bmatrix}x&-y\\[0.3em]-z&w\end{bmatrix}\)+ \(\begin{bmatrix}-2x&0\\[0.3em]0&-2w\end{bmatrix}\) = \(\begin{bmatrix}x+x+(-2y)&y+(-y)+0\\[0.3em]z+(-z)+0&w+w+(-2w)\end{bmatrix}\) = \(\begin{bmatrix}0&0\\[0.3em]0&0\end{bmatrix}\)

6264.

If A = \(\begin{bmatrix}x&y\\[0.3em]z&w\end{bmatrix},\) B = \(\begin{bmatrix}x&-y\\[0.3em]-z&w\end{bmatrix}\) and C = \(\begin{bmatrix}-2x&0\\[0.3em]0&-2w\end{bmatrix},\) then A + B + C is a(a) identity matrix (b) null matrix (c) row matrix (d) column matrix

Answer»

(b) A + B + C = \(\begin{bmatrix}x&y\\[0.3em]z&w\end{bmatrix}\)+ \(\begin{bmatrix}x&-y\\[0.3em]-z&w\end{bmatrix}\)+ \(\begin{bmatrix}-2x&0\\[0.3em]0&-2w\end{bmatrix}\)

= \(\begin{bmatrix}x+x+(-2y)&y+(-y)+0\\[0.3em]z+(-z)+0&w+w+(-2w)\end{bmatrix}\)

= \(\begin{bmatrix}0&0\\[0.3em]0&0\end{bmatrix}\)

Discussion

6265.	Which of the following statement is incorrect in reference to compost?1. In this, biodegradable waste is decomposed in pits.2. It is rich in organic matter and nutrients3. The process of decomposition in pits is slowed down by earthworms4. The waste material decomposed is farm waste material like livestock excreta, vegetable waste etc.
Answer» Correct Answer - Option 3 : The process of decomposition in pits is slowed down by earthworms The process of decomposition in pits is slowed down by earthworms is an incorrect statement. The process of decomposition in pits is fastened by earthworms not slowed down. Decomposition is the process by which dead organic substances are broken down into simpler organic or inorganic matter such as carbon dioxide, water, simple sugars and mineral salts. Compost is made by decomposing organic materials into simpler organic and inorganic compounds by the microorganisms in a process called composting.

6265.

Which of the following statement is incorrect in reference to compost?1. In this, biodegradable waste is decomposed in pits.2. It is rich in organic matter and nutrients3. The process of decomposition in pits is slowed down by earthworms4. The waste material decomposed is farm waste material like livestock excreta, vegetable waste etc.

Answer» Correct Answer - Option 3 : The process of decomposition in pits is slowed down by earthworms

The process of decomposition in pits is slowed down by earthworms is an incorrect statement.

The process of decomposition in pits is fastened by earthworms not slowed down.

Decomposition is the process by which dead organic substances are broken down into simpler organic or inorganic matter such as carbon dioxide, water, simple sugars and mineral salts.
Compost is made by decomposing organic materials into simpler organic and inorganic compounds by the microorganisms in a process called composting.

Discussion

6266.	At what temperature will the water density be maximum?A. 0°CB. 4°CC. 39°CD. 100°C1. A2. B3. D4. C
Answer» Correct Answer - Option 2 : B The correct answer is 4°C. At 4°C temperature, the water density will be maximum. At 0°C temperature, the water density will be the minimum. The boiling point of water is 100°C. The cluster starts forming at 4-degree Celcius. The molecules are still slowing down and coming closer together but the formation of clusters makes the molecules be further apart. The density starts to decrease because the cluster formation is a bigger effect.

6266.

At what temperature will the water density be maximum?A. 0°CB. 4°CC. 39°CD. 100°C1. A2. B3. D4. C

Answer» Correct Answer - Option 2 : B

The correct answer is 4°C.

At 4°C temperature, the water density will be maximum.
At 0°C temperature, the water density will be the minimum.
The boiling point of water is 100°C.
The cluster starts forming at 4-degree Celcius.
The molecules are still slowing down and coming closer together but the formation of clusters makes the molecules be further apart.
The density starts to decrease because the cluster formation is a bigger effect.

Discussion

6267.	If `emf` of battery is `100V`, then what was the resistance of Rheostat adjusted at reading (`i=2A`, `V=20V`) A. `10 Omega`B. `20 Omega`C. `30 Omega`D. `40 Omega`
Answer» Correct Answer - D from the curve slope `=(1)/(v)=(1)/(R )=(1)/(10)` , `R=10Omega` for second reading `i=(Emf)/(R+R_(rh)implies 2=(100)/(10+R_(rh))impliesR_(rh)=40Omega`

Discussion

6268.	If `emf` of battery is `100V`, then what was the resistance of Rheostat adjusted at reading (`i=2A`, `V=20V`) A. `10Omega`B. `20Omega`C. `30Omega`D. `40Omega`
Answer» From the curve slope `= (I)/(v) = (1)/(R) = (1)/(10) R = 10Omega` for second reading `I = (Emf)/(R+R_(rh)) 2 = (100)/(10 + R_(rh)) rArr R_(rh) = 40Omega` .

Discussion

6269.	If a tuning fork of frequency `(340 pm 1 %)` is used in the resonance tube method and the first and second resonance lengths are `20.0cm` and `74.0 cm` respectively. Find the maximum possible percentage error in speed of sound.A. `5.03m//s`B. `0.503m//s`C. `2.51m//s`D. `0.251m//s`
Answer» Correct Answer - A `l_(1)=20.0cm to Deltal_(1)=0.1 cm` `l_(2)=74.0cm to Deltal_(2)=0.1 cm` `{:(f_(0)=(340HZ+-1%),(Deltaf_(0))/(f_(0))=1%=(1)/(100)):} ` `((dv)/(v))_(max)=(Deltaf_(0))/(f_(0))+(Deltal_(2)+Deltal_(1))/(l_(2)-l_(1))`

Discussion

6270.	The e.m.f. induced in a coil of wire, which is rotating in a magnetic field, does not depend onA. number of turns in coilB. resistance of coilC. rate of change of fluxD. All of these
Answer» Correct Answer - B Induced emf in a coil is independent of the resistance of coil.

Discussion

6271.	An object is placed infront of a convex mirror of focal length f. Find the maximum and minimum distance of two object from the mirror such that the image is real and magnified.A. `20 and oo`B. f and 2fC. f and 0D. None of these
Answer» Correct Answer - B For real image v = 0 `therefore ` From `(1)/(f) = (1)/(v) =(1)/(u)`

Discussion

6272.	Taking the open end of tube as `y =0` position of pressure nodes will be .A. `y = - 1cm y = 49cm`B. `y = 0 cm y = 50cm`C. `y = 1cm y = 51cm`D. None of these
Answer» `y = - 1 cm, y = 49cm` .

Discussion

6273.	If a tuning fork of frequency `(340 pm 1 %)` is used in the resonance tube method and the first and second resonance lengths are `20.0cm` and `74.0 cm` respectively. Find the maximum possible percentage error in speed of sound.
Answer» `l_(1) = 20.0 cm rarr Deltal_(1) = 0.1 cm` `l_(2) = 74.0 cm rarr Deltal_(2) = 0.1 cm` `f_(0) = 9340 Hz -+ 1%) (Deltaf_(0))/(f_(0)) = 1% = (1)/(100)` `((DeltaV)/(V))_(max) = (Deltaf_(0))/(f_(0)) + (Deltal_(2)+Deltal_(1))/(l_(2)-l_(1)) = (1)/(100) + (0.1 = 0.1)/(74.0 - 24.0) = (1)/(100) + (0.2)/(50.0) = 0.014` .

Discussion

6274.	The average power dissipation in a pure capacitance in `AC` circuit isA. `(1)/(2)CV^(2)`B. `CV^(2)`C. `(1)/(4)CV^(2)`D. zero
Answer» Correct Answer - D In purely inductive or purely capacitive circuit, power loss is zero.

Discussion

6275.	Two infinitely long rods carry equal linear denisty `lambda` each. They are perpendicular to each other and they are in different planes and separated by a distance `d`. Find the electrostatic force on once rod due to the other
Answer» Correct Answer - A::B::D Two infinetely long rods carry equal linear density `lamda` each. They are perpendicular to each other and they are in different planes and separated by a distance d. Find the electrostatic force on one rod due to the other,

Discussion

6276.	Three containers of same base area, same height are filled with three different liquids of same mass as shown in the figure. If `F_(1),F_(2),F_(3)` are the force exerted by the liquid on the base of the container in case I, and II respectively, then we have the relation: A. `F_(1) = F_(2) = F_(3)`B. `F_(1) gt F_(2) gt F_(3)`C. `F_(3) gt F_(2) gt F_(1)`D. `F_(2) gt F_(3) gt F_(1)`
Answer» Correct Answer - D Force on bottom surface `= rho gH xx A`

Discussion

6277.	To activate the reaction `(n, alpha)` with stationary `B^(11)` nuclei, neutrons must have the activation kinetic energy `T_(th) = 4.0 MeV(n, alpha)` means that `n` is bombarded to obtain `alpha`. Find the energy of this reaction.
Answer» `T_("in") =-(m_(T)+m_(P))/(m_(T)).Q` `rArr Q = -T_("in")(m_(T))/(m_(T)+m_(P))` `=-4MeV xx (11)/(11+1) = -3.7 MeV`.

Discussion

6278.	Why is the belt of a van de Graaff generator made of insulating material?
Answer» Due to this charges taken by the belt do not spread out but they are stored Charges are send by the belt to the pointed conductor. If the belt is conductor material the charges spread out on the whole belt and the work system does not run systematically.

Discussion

6279.	Assertion: Bulb generally get fused when they are switched on or off. Reason: When we switch on or off, a circuit current changes in it rapidly.
Answer» Correct Answer - D Switching results in high decay/growth rate of current which results in a high. Current when bulb is turned off (due to back emf). So, a bulb is must likely to get fused when it is just turned off.

Discussion

6280.	(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Answer» (a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field. (b) Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude. (c) An electron traveling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left-hand rule, magnetic field should be applied in a vertically downward direction.

6280.

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Answer»

(a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.

(b) Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.

(c) An electron traveling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain undeflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left-hand rule, magnetic field should be applied in a vertically downward direction.

Discussion

6281.	A circuit contains an ideal battery, three resistors and two ideal ammeters. The ammeters read 0.2 amp and 0.3 amp. After two of the resistor are switched (inerchanged) the readings of ammeter do not change. Then A. If `R_(2)=R_(1), I_(1)=I_(2)=0.15` amp and `I_(3)=0.05` ampB. If `R_(2)=R_(1),` the battery current is `I=0.20` ampC. If `R_(2)=R_(3), I_(2)=I_(3)=0.10` amp and `I_(1)=0.20` ampD. If `R_(2)=R_(3)`, the battery current is `I=0.40` amp
Answer» Correct Answer - A::C::D As the ammeters are ideal, so `R_(eq)=(R_(1)R_(2)R_(3))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))` The readings of two ammeters are different so resistors `R_(1)` and `R_(3)` are different. If `R_(1)` and `R_(3)` are interchanged. The reading of ammeters will also change. If `R_(2)=R_(1)` (if `R_(2)` and `R_(1)` switched the reading of the ammeter do not vary) similarly if `R_(2)=R_(3)` (`R_(2)` and `R_(3)` switched the reading oif ammeters do not change) If `R_(2)=R_(1),I_(1)=I_(2)=0.15, I_(3)=0.05` and `I=0.35` If `R_(2)=R_(3), I_(2)=I_(3)=0.10, I_(1)=0.20` and `I=0.40` amp

Discussion

6282.	Consider the circuit consisting of two capacitors having capacitances `3 mu F` and `6 mu F` and an ideal battery of emf e = 4 volts as shown. The switch S is open for a long time and then closed. After the switch is closed. The work done by the battery is : A. 16 m JB. 32 m JC. 64 m JD. 128 m J
Answer» Correct Answer - C Initial charge on 6 mF capacitor `= (3 xx 6)/(3+6) xx 4 = 8 mC` final charge on 6mF capacitor `= 6 xx 6 = 25 mC` charge passing through cell (in direction aided by cell) `= q_(f) -q_(i) = Dq = 24 - 8 = 16 mC` work done by cell = Dqe `= 16 xx 4 = 64 mJ`

Discussion

6283.	A meter bridge is set up as shown, to determine an unknown resistance `X` using a standard 10 ohm resistor. The galvanometer shows null point when tapping -key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determine value of `X` is A. `10.2ohm`B. `10.6ohm`C. `10.8ohm`D. `11.1ohm`
Answer» `l_(1) = 52 + 1 = 53 cm` `l_(2) = 48 + 2 = 50 cm` `(l_(1))/(l_(2)) = (x)/(R) rArr (53)/(50) = (x)/(10)` `x = 10.6 Omega` .

Discussion

6284.	A meter bridge is set up as shown, to determine an unknown resistance `X` using a standard 10 ohm resistor. The galvanometer shows null point when tapping -key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determine value of `X` is A. 10.2 ohmB. 10.6 ohmC. 10.8 ohmD. 11.1 ohm
Answer» Correct Answer - B `l_(1) = 52 +1 = 53 cm` `l_(2) = 48 + 2 =50 cm` `(l_(1))/(l_(2))= (x)/(R)`.

Discussion

6285.	The power of a motor pump is 2 kw.How much water per minute the pump can raise to a height of 10m?(GIVEN G=10m/s)
Answer» Given, power of pump = 2kW =2000W Time (t)= 60sec Height (h) = 10m g = 10m/s² As you know, power =work done per unit time. Work done = mgh = m ×10×10 =100m Therefore, 2000W = 100m/60s Therefore, m = 1200kg So, the pump can raise 1200kg of water in one minute. 2 kW is equal to 2 x 1000 W (as 1 kW = 1000 W). work = mgh, where m = mass , g=acceleration due to gravity( usually 9.8 or 10 ) and h= height. Power = work/time work done = m x 10 x 10. work done = m x 100 J (joule). power = J/sec (work/time). 2000 = 100m/60 as 1 minute is equal to 60 sec. 2000 x 60 = 100m 120000/100=m m=1200 kg 1200 kg = 1200 liter ( here water is being raised so in water 1 kg will be equal to 1 liter)

6285.

The power of a motor pump is 2 kw.How much water per minute the pump can raise to a height of 10m?(GIVEN G=10m/s)

Answer»

Given, power of pump = 2kW =2000W

Time (t)= 60sec

Height (h) = 10m

g = 10m/s²

As you know, power =work done per unit time.

Work done = mgh = m ×10×10 =100m

Therefore, 2000W = 100m/60s

Therefore, m = 1200kg

So, the pump can raise 1200kg of water in one minute.

2 kW is equal to 2 x 1000 W (as 1 kW = 1000 W).
work = mgh, where m = mass , g=acceleration due to gravity( usually 9.8 or 10 ) and h= height.
Power = work/time
work done = m x 10 x 10.
work done = m x 100 J (joule).
power = J/sec (work/time).
2000 = 100m/60 as 1 minute is equal to 60 sec.
2000 x 60 = 100m
120000/100=m
m=1200 kg
1200 kg = 1200 liter ( here water is being raised so in water 1 kg will be equal to 1 liter)

Discussion

6286.	In an electromagnetic wave, the electric field oscillates sinusoidally with amplitude `48 Vm^(-1)`, the RMS value of oscillating magnetic field will be nearly equal to :A. `1.6 xx 10^(-8) T`B. `16 xx 10^(-9) T`C. `144 xx 10^(8) T`D. `11.3 xx 10^(-8) T`
Answer» Correct Answer - D `B_(0)=(E_(0))/(c)=(48)/(3xx10^(8))=16 xx10^(-8)T` `B_("rms") =(B_(0))/(sqrt(2))=(16 xx10^(-8))/(sqrt(2))=8 sqrt(2) xx 10^(-8) T`.

Discussion

6287.	One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will ……..(a) get shifted downwards (b) get shifted upward (c) will remain the same (d) data insufficient to conclude
Answer» (b) get shifted upward

Discussion

6288.	A diode detector is used to detect an amplitude modulated wave of `60%` modulation by using a condense of capacity `250` picofarad in parallel with a load resistance `100` kilo ohm find the maximum modulated which could be find the maximum modulated frequency which could be detected by itA. `10.32 Mhz`B. `10.61 kHz`C. `5.31 MHz`D. 5.31 kHz
Answer» Correct Answer - B `therefore tau=RC = 100 xx 10^(3) xx 250 xx 10^(-12)`s `=2.5 xx 10^(7) xx 10^(-12) s=2.5 xx 10^(-6)`s The higher frequency which can be detected with tolerable distortion is `f=1/(2pim_(a)RC)= 1/(2pixx0.6xx2.5xx10^(-6))`Hz `=(100 xx 10^(4))/(25 xx 12pi)`Hz `=(4/12pi xx 10^(4)` Hz = 10.61 kHz

Discussion

6289.	______ is the room where you relax, watch television. A) living roomB) bathroom C) kitchen D) laundry room
Answer» Correct option is A) living room

Discussion

6290.	Ques. A metallic sphere of radius r remote from all other bodies is irradiated with a radiation of wavelength  which is capable of causing photoelectric effect. (A) the maximum potential gained by the sphere will be independent of its radius (B) the net positive charge appearing on the sphere after a long time will depend on the radius of the sphere (C) the kinetic energy of the most energetic electrons emanating from the sphere will keep on declining with time (D) the kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the spher
Answer» Correct answer is (A) The maximum potential gained by the sphere will be independent of its radius The maximum potential gained by the sphere will be independent of its radius. The net positive charge appearing on the sphere after a long time will depend on the radius of the sphere. The kinetic energy of the most energetic electrons emanating from the sphere will keep on declining with time. The kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the sphere.

6290.

Ques. A metallic sphere of radius r remote from all other bodies is irradiated with a radiation of wavelength  which is capable of causing photoelectric effect. (A) the maximum potential gained by the sphere will be independent of its radius (B) the net positive charge appearing on the sphere after a long time will depend on the radius of the sphere (C) the kinetic energy of the most energetic electrons emanating from the sphere will keep on declining with time (D) the kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the spher

Answer»

Correct answer is (A) The maximum potential gained by the sphere will be independent of its radius

The maximum potential gained by the sphere will be independent of its radius. The net positive charge appearing on the sphere after a long time will depend on the radius of the sphere.

The kinetic energy of the most energetic electrons emanating from the sphere will keep on declining with time. The kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the sphere.

Discussion

6291.	______ is used to control something like television from a distance. A) remote control B) camera C) telescope D) microphone
Answer» Correct option is A) remote control

Discussion

6292.	Which of the following is not equal to watt?A. Joule/secondB. Ampere`xx`voltC. (Ampere`)^2xx`ohmD. Ampere/volt
Answer» Correct Answer - D Watt=joule/second=ampere`xx`volt=ampere`^2``xx`ohm

Discussion

6293.	Which of the following quantities has its unit as newton - second?A. VelocityB. Angular momentumC. MomentumD. Energy
Answer» Correct Answer - C impulse`=`chage in momentum `=Fxxt` So the unit of mementum will be equal to newton-sec.

Discussion

6294.	A ball moves over a fixed track as shown in the figure. From `A` to `B` ball rolls without slipping. Surface `BC` is frictionless. `K_(A), K_(B)` and `K_(C)` are kinetix energies of the ball at `A, B` and `C`, respectively. Then .A. a,bB. a,cC. b,dD. None of these
Answer» Correct Answer - A `TE._(A)=T.E_(B)=T.E_(C)` `0+M"g"h_(A)=1/2mv^(2)+1/2Iomega^(2)+m"g"r` `" "=1/2I omega^(2)+M"g"hC` `"So "k_(B)gtk_(C)gtk_(A)` `"So "h_(A)gth_(c)`

Discussion

6295.	An insect of negligible mass is sitting on a block of mass M, tied with a spring of force constant k. the block performs simple harmonic motion with amplitude A in front of a plane mirror placed as shown in the figure the maximum speed of insect relative to its image will beA. `Asqrt((k)/(m))`B. `(Asqrt(3))/(2)sqrt((k)/(m))`C. `Asqrt(3)sqrt((k)/(m))`D. `Asqrt((m)/(k))`
Answer» Correct Answer - C `v_(max)=Aomega`

Discussion

6296.	An insect of negligible mass is sitting on a block of mass M, tied with a spring of force constant K. The block performs simple harmonic motion with amplitude A infront of a plane mirror placed as shown. The maximum speed of insect relative to its image will be A. `A sqrt(3) sqrt(k/M)`B. `Asqrt(k/M)`C. `(A sqrt(3))/2 sqrt(k/M)`D. `2A sqrt(k/M)`
Answer» Correct Answer - C

Discussion

6297.	A sensor is exposed for time t to a lamp of power P placed at a distance l. The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is `lambda` is:A. `N= (4P lambda d^(2) t)/(hc l^(2))`B. `N=(P lambda d^(2) t)/(4h c l^(2))`C. `N=(P lambda d^(2) t)/(16 hc l^(2))`D. `N=(P lambda d^(2) t)/(hc l^(2))`
Answer» Correct Answer - D

Discussion

6298.	A sensor is exposed for 0.1 s to a 200 W lamp 10m away The sensor has an opening that is 20 mm in diameter. How many photons enter the sensor if the wavelength of light is 600nm? Assume that all the energy of the lamp is given off as light.
Answer» Correct Answer - 1 Number of photons entering sensor are `N=(P_(i) lambda t)/(hc)" "( P_(i) rarr" incident power")` Where `P_(i)=P/(4pi l^(2))xx pi (2d)^(2)` `implies N=(P lambda d^(2) t)/(hcl^(2))`

Discussion

6299.	Consider the uniform magnetic field shown: Starting from point P and without leaving the region of magnetic field, is it possible to choose a closed path (that is, a path that returns to P) for which the line integral of the magnetic field is nonzero?A. Yes, but only positiveB. Yes, but only negativeC. Yes, both positive and negativeD. No.
Answer» Correct Answer - D From ampere circuital law, not current threading the amperian kloop will be zero.

Discussion

6300.	A particle of mass m is projected form ground with velocity u making angle `theta` with the vertical. The de Broglie wavelength of the particle at the highest point isA. `infty`B. `(h)/(m u sin theta)`C. `(h)/(m u cos theta)`D. `(h)/(m u)`
Answer» Correct Answer - B Velocity at highest point `= u sin theta` `:. lamda_(D) = (h)/(m u sin theta)` (since `theta` is velocity w.r.t. vertical)

Discussion

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