100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 1957 InterviewSolution

97801.	Assertion`:-` The displacement always equals the product of the average velocity and the time interval. Reason`:-` The average velocity always equals the mean value of the initial and final velocities.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False
Answer» Correct Answer - C

Discussion

97802.	Assertion`:-` The cord between `A` and `B` more likely to snap with a gradual downward pull at `D`. Reason`:-` When someone gives the lower cord a sudden downward pull, the tension in the lower cord inceaces suddenly.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False
Answer» Correct Answer - B

Discussion

97803.	What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of `30^(@)` with the direction of a uniform magnetic field of 0.20 T ?A. 0.4 N/mB. 0.6 N/mC. 0.8 N/mD. 0.10 N/m
Answer» Correct Answer - C `vecF=I vecl xx vecB` `F= I l B sin theta ` `(F)/( l) = 8 xx (0.2) sin 30^(@) = 0.8 N//m`

Discussion

97804.	A cord is used to lower vertically a block of mass `M`, a distance `d` at a constant downward acceleration of `(g)/(4)`, then the work done by the cord on the block isA. `mg d//4`B. `3 Mg d//4`C. `-3 Mg`D. `mg d`
Answer» Correct Answer - C `Mg - T = Ma = (Mg)/(4)` `T = (3 Mg)/(4)` `W = - T xx d = (-3 Mgd)/(4)`

Discussion

97805.	What is line current?
Answer» The current flowing between any two phases of the winding is called line current and it is denoted by the letter I_L.

Discussion

97806.	What is balanced load and unbalance load?
Answer» Balanced load: In a three-phase system the power factors and the phase current or line currents of the 3-phase are equal, then that load is called balanced load. Unbalance load: If the three-phases have different power factors and the phase current, then the load is called the unbalance load.

97806.

What is balanced load and unbalance load?

Answer»

Balanced load:

In a three-phase system the power factors and the phase current or line currents of the 3-phase are equal, then that load is called balanced load.

Unbalance load:

If the three-phases have different power factors and the phase current, then the load is called the unbalance load.

Discussion

97807.	The current (I) and voltage (V) graphs for a given metallic wire at two different temperature `(T_(1))` and `(T_(2))` are shown in fig. It is concluded that A. `T_(1) lt T_(2)`B. `T_(1) gt T_(2)`C. `T_(1) = T_(2)`D. `T_(1) = 2T_(2)`
Answer» Correct Answer - B Slope of graph `V-I = (1)/("Resistance")` `R_(2) gt R_(1)` So` T_(2) gt T_(1)`

Discussion

97808.	What are the advantages of high voltage transmission?
Answer» The advantages of high voltage transmission. a. Saving in conductor materials. b. Low power loss (I² R) of transmission lines due to decrease in current. c. Better efficiency of line due to fewer losses. d. Better voltage regulation due to less voltage drop in line due to less transmission current. e. Due to the less cross-section of conductor distance between the poles increases and the cost decreases and the labour cost also decreases.

97808.

What are the advantages of high voltage transmission?

Answer»

The advantages of high voltage transmission.

a. Saving in conductor materials.

b. Low power loss (I² R) of transmission lines due to decrease in current.

c. Better efficiency of line due to fewer losses.

d. Better voltage regulation due to less voltage drop in line due to less transmission current.

e. Due to the less cross-section of conductor distance between the poles increases and the cost decreases and the labour cost also decreases.

Discussion

97809.	What are the advantages of electro magnetism?
Answer» The advantages of electromagnetism are given below. a. Electro magnets can be magnetised very easily by sending DC through it. b. Changing the direction of the current through the coil can change the polarity of the poles. c. The strength of the magnet can be controlled by the electric current. d. Electro magnets can be made in any shape depending upon the need. e. The magnetic strength remains constant as long as the current is constant.

97809.

What are the advantages of electro magnetism?

Answer»

The advantages of electromagnetism are given below.

a. Electro magnets can be magnetised very easily by sending DC through it.

b. Changing the direction of the current through the coil can change the polarity of the poles.

c. The strength of the magnet can be controlled by the electric current.

d. Electro magnets can be made in any shape depending upon the need.

e. The magnetic strength remains constant as long as the current is constant.

Discussion

97810.	The ratio of tensions in the string connected to the block of mass `m_(2)` in figure, respectively, is (friction is absent everywhere):`[m_(1)=50 kg, m_(2)=80kg` and `F=1000N]` A. `7:2`B. `2:7`C. `3:4`D. `4:3`
Answer» Correct Answer - 3 In fig. (a)`1000-T_(1)=50a ` in fig(b) `T_(1)-800=80a` `1000+500-T_(2)=50a` `T_(1)=12,000/13 " " T_(2)-800=80a` `T_(2)=16,000/13`

Discussion

97811.	Read the following statements carefully : Y : The resistivity of a semiconductor decreases with increase of temperature. Z : In a conducing solid, the rate of collisions between free electrons and ions increase with increase of temperature. Select the correct statement (s) from the followin :A. Y is true, but Z is falseB. Y is false, but Z is trueC. Both Y and Z are trueD. Y is true, and Z is the correct reason for Y
Answer» Correct Answer - C c. Both are basic facts.

Discussion

97812.	Cylinderical lenses are used to correct the eye defect called-(A) myopia(B) hypermetropia(C) astrigmatism(D) Presbyopia
Answer» (C) astrigmatism

Discussion

97813.	The difference in tensions in the string at lowest and highest points in the path of the particle of mass ‘m’ performing vertical circular motion is: (A) 2 mg (B) 4 mg (C) 6 mg (D) 8 mg
Answer» Answer is: (C) 6 mg

97813.

The difference in tensions in the string at lowest and highest points in the path of the particle of mass ‘m’ performing vertical circular motion is: (A) 2 mg (B) 4 mg (C) 6 mg (D) 8 mg

Answer»

Answer is:

(C) 6 mg

Discussion

97814.	With the increase of temperature the resistivity of semiconductor-(A) increases (B) decreases(C) remains constant(D) becomes Zero
Answer» (B) decreases

Discussion

97815.	Let U1 ≡ a1x + b1y + c1 = 0, U2 ≡ a2x + b2y + c2 = 0 and U3 = a3x + b3y + c3 = 0 be three lines such that no two are parallel. If there exists non-zero real numbers λ1, λ2 and λ3 such that λ1U1 + λ2U2 + λ3U3 = 0, then the three lines U1 = 0, U2 = 0 and U3 = 0 are concurrent.
Answer» Suppose λ₁U₁ + λ₂U₂ + λ₃U₃ = 0,where λ₁, λ₂ and λ₃ K3 are non-zero real numbers. Therefore U₃ = (-λ₁/λ₃) + (-λ₂/λ₃)U₂ which is of the form λ₁U₁ + μU₂ = 0. Hence, U₃ = 0 passes through the point of intersection of the lines U₁= 0 and U₂ = 0. Therefore, the three lines are concurrent.

97815.

Let U1 ≡ a1x + b1y + c1 = 0, U2 ≡ a2x + b2y + c2 = 0 and U3 = a3x + b3y + c3 = 0 be three lines such that no two are parallel. If there exists non-zero real numbers λ1, λ2 and λ3 such that λ1U1 + λ2U2 + λ3U3 = 0, then the three lines U1 = 0, U2 = 0 and U3 = 0 are concurrent.

Answer»

Suppose λ₁U₁ + λ₂U₂ + λ₃U₃ = 0,where λ₁, λ₂ and λ₃ K3 are non-zero real numbers. Therefore

U₃ = (-λ₁/λ₃) + (-λ₂/λ₃)U₂

which is of the form λ₁U₁ + μU₂ = 0. Hence, U₃ = 0 passes through the point of intersection of the lines U₁= 0 and U₂ = 0. Therefore, the three lines are concurrent.

Discussion

97816.	When an unpolarized light of intensity `I_(0)` is incident on a polarizing sheet, the intensity of the light which does not get transmitted isA. `I_(0)`B. 0C. `I_(0)//2`D. `I_(0) cos^(2)theta`
Answer» Correct Answer - C `I_(0)` (unpolarised ) `rarr(I_(0))/(2)`

Discussion

97817.	Frame and solve the node equations of the network of Fig. Hence, find the total power consumed by the passive elements of the network.
Answer» The node equation for node 1 is V₁(1 + 1 + 1/0.5) - V₂/0.5 - 15/1 = 0 or 4V₁ - 2V₂ = 15 Similarly, for node 2, we have V₁(1 + 1/2 + 1/0.5) - V₂/0.5 - 20/1 = 0 or 4V₁ − 7V₂ = − 40 ...(ii) ∴ V₂ = 11 volt and V₁ = 37/4 volt Now, I₁ = (15 - 37/4)/I = 23/4A = 5.75A; I₂ = (11 - 37/4)/0.5 = 3.5A I₄ = 5.75 + 3.5 = 9.25A, I₃ = (20 - 11)/1 = 9A; I₅ = 9 - 3.5 = 5.5A The passive elements of the network are its five resistances. Total power consumed by them is = 5.75² × 1 + 3.5² × 0.5 + 9² × 1 + 9.25² × 1 + 5.5² × 2 = 266.25

97817.

Frame and solve the node equations of the network of Fig. Hence, find the total power consumed by the passive elements of the network.

Answer»

The node equation for node 1 is

V₁(1 + 1 + 1/0.5) - V₂/0.5 - 15/1 = 0

or 4V₁ - 2V₂ = 15

Similarly, for node 2, we have

V₁(1 + 1/2 + 1/0.5) - V₂/0.5 - 20/1 = 0

or 4V₁ − 7V₂ = − 40 ...(ii)

∴ V₂ = 11 volt and V₁ = 37/4 volt

Now,

I₁ = (15 - 37/4)/I = 23/4A = 5.75A; I₂ = (11 - 37/4)/0.5 = 3.5A

I₄ = 5.75 + 3.5 = 9.25A, I₃ = (20 - 11)/1 = 9A; I₅ = 9 - 3.5 = 5.5A

The passive elements of the network are its five resistances. Total power consumed by them is

= 5.75² × 1 + 3.5² × 0.5 + 9² × 1 + 9.25² × 1 + 5.5² × 2 = 266.25

Discussion

97818.	Applying Kirchhoff's laws to different loops in Fig. find the values of V1 and V2.
Answer» Starting from point A and applying Kirchhoff's voltage law to loop No.3, we get - V₃+ 5 = 0 or V₃ = 5 V Starting from point A and applying Kirchhoff's voltage law to loop No.1, we get 10 - 30 - V, + 5 =0 or V, =-15 V The negative sign of V, denotes that its polarity is opposite to that shown in the figure. Starting from point B in loop No.3, we get -(-15) - V₂ + (-15) = 0 or V₂= 0

97818.

Applying Kirchhoff's laws to different loops in Fig. find the values of V1 and V2.

Answer»

Starting from point A and applying Kirchhoff's voltage law to loop No.3, we get

- V₃+ 5 = 0 or V₃ = 5 V

Starting from point A and applying Kirchhoff's voltage law to loop No.1, we get

10 - 30 - V, + 5 =0 or V, =-15 V

The negative sign of V, denotes that its polarity is opposite to that shown in the figure.

Starting from point B in loop No.3, we get

-(-15) - V₂ + (-15) = 0 or V₂= 0

Discussion

97819.	IF U=3x2y,V=xz2 -2y Evaluate [(grad U) (grad V)]
Answer» U = 3x²y grad U = ∇ U = (i ∂/∂x + j ∂/∂y + k ∂/∂z) 3x²y = 6xy i + 3x²j Also, V = xz² - zy grad V = ∇ V = (i ∂/∂x + j ∂/∂y + k ∂/∂z) (xz² - 2y) = z² i - 2 j + 2xz k Now, (grad U).(grad V) = (6xy i + 3x² j + 0 k) (z² i - 2 j + 2xz k) = 6xyz²+ 6x² = 6x (yz² + x)

97819.

IF U=3x2y,V=xz2 -2y Evaluate [(grad U) (grad V)]

Answer»

U = 3x²y

grad U = ∇ U = (i ∂/∂x + j ∂/∂y + k ∂/∂z) 3x²y

= 6xy i + 3x²j

Also, V = xz² - zy

grad V = ∇ V = (i ∂/∂x + j ∂/∂y + k ∂/∂z) (xz² - 2y)

= z² i - 2 j + 2xz k

Now, (grad U).(grad V) = (6xy i + 3x² j + 0 k) (z² i - 2 j + 2xz k)

= 6xyz²+ 6x²

= 6x (yz² + x)

Discussion

97820.	Determine current through 6 ohm resistance connected across A-B terminals in the electric circuit of Fig. using Thevenin’s Theorem.
Answer» Applying Thevenin’s theorem, after detaching the 6-ohm resitor from terminals A −B, V_TH= V_C = 15 − 1 × 3 = 12 volts R_TH= 4 + 3/6 = 6 ohms I_L = 12/(6 + 6) = lamp

97820.

Determine current through 6 ohm resistance connected across A-B terminals in the electric circuit of Fig. using Thevenin’s Theorem.

Answer»

Applying Thevenin’s theorem, after detaching the 6-ohm resitor from terminals A −B,

V_TH= V_C = 15 − 1 × 3 = 12 volts

R_TH= 4 + 3/6 = 6 ohms

I_L = 12/(6 + 6) = lamp

Discussion

97821.	The electric field intensity of an infinite long charge line varies
Answer» the electric field intensity of an infinite long charge line varies with distance.

Discussion

97822.	A rectangular coil `20 cm xx 10 cm` having 500 turns rotates in a magnetic field of `5 xx 10^(-3) T` with a frequency of `1200 rev min^(-1)` about an axis perpendicular to the field.A. The maximum value of the induced emf `2pi//5` volt.B. The instantaneous emf when the plane of the coil is perpendicular to the field is zeroC. The instantaneous emf when the plane of the coil makes an angle of `60^(@)` with the field is `pi//5` volt.D. The instantaneous emf when the plane of the coil makes an angle of `30^(@)` with the field is `pi//10` volt
Answer» Correct Answer - A::B::C `E=NAB omega cos theta`, where `theta` is the angle between plane of coil and magnitude field, Substituting values option (a), (b) and (c) are correct.

Discussion

97823.	A square conducting loop is placed in the neighbourhood of a coplaner long straight wire carrying a current i. A. if `(di)/(dt)=0`, no current is induced in the loopB. if `(di)/(dt)gt0` , current in the loop is clockwise.C. if `(di)/(dt)lt0` , current in the loop is anticlockwise.D. if `(di)/(dt)gt0` , current in the loop is anticlockwise.
Answer» Correct Answer - A::D If `(di)/(dt)=0`, it means current is constant, its flux through loop with remain constant. Hence no emf is induced. If `(di)/(dt)gt0`, current increases so flux increates in loop in inward direction, so induced current should be anticlockwise which produces flux in outward direction.

Discussion

97824.	Consider a variation of the previous problem. Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistance but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity `(omega)` in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle `(theta)` with the vertical.A. `(B^(2)a^(3)omega)/(2R)-mg sin theta`B. `(B^(2)a^(3)omega)/(2R)+mg sin theta`C. `(B^(2)a^(3)omega)/(R)-mg sin theta`D. None of these
Answer» Correct Answer - A We know that `F=(B^(2)a^(2)omega)/(2R)=0Bl` Component of mg along `F=mg sin theta` Net force `=(B^(2)a^(2)omega)/(2R)-mg sin theta`.

Discussion

97825.	Figure shows a situation similar to the previous problem. All parameters are the same except that a battery of emf `epsilon` and a variable resistance R are connected between O and C. The connecting wires have zero resistance. No external force is applied on the rod (except gravity, forces by the magnetic field and by the pivot). In what way should the resistance R be changed so that the rod may rotate with uniform angular velocity in the clockwise direction? Express your answer in terms of the given quantities and the angle `(theta)` made by the rod OA with the horizontal.A. `(Bomega a^(2)-2epsilon)(aB)/(2 mg cos theta)`B. `(Bomega a^(2)+2epsilon)(aB)/(mg cos theta)`C. `(Bomega a^(2)+2epsilon)(aB)/(2 mg cos theta)`D. None of these
Answer» Correct Answer - C `Emf=(Bomega a^(2))/(2)` [from previous problem]` Current =(e + epsilon)/(R)=(Bomega a^(2)+epsilon)/(2R)=(Bomega a^(2)+2epsilon)/(2R)` `implies mg cos theta = ilB` [net force acting on the rod is 0] `mg cos theta=((Bomega a^(2)+2 epsilon)aB)/(2R)(axxB)`.

Discussion

97826.	The half lifr of a particle of mass `1.6 xx 10^(-26) kg` is `6.9 s` and a stream of such particles is travelling with the kinetic energy of a particles is travelling with the kinetic energy of a particle being 0.05 eV. The fraction of particles which will decay when they travel a distance of 1 m isA. `0.1`B. `0.01`C. `0.001`D. `0.0001`
Answer» Correct Answer - D `(dN)/(dt)=n lamda rArr(dN)/(dt)=lamda dt` `(6.63)/(6.9)xx10^(-4) =10^(4)`

Discussion

97827.	I don’t feel like swimming. I’d rather sun-bathe on the ________. A) beach B) coast C) dust D) harbor E) shore
Answer» Correct option is A) beach

Discussion

97828.	Which element has highest Electron Affinity ?A. BeB. BC. LiD. C
Answer» Correct Answer - D On moving left to right in a period electron affinity increases.

Discussion

97829.	Amount of `80%` pure NaOH sample which is required to completely react with `42.6 gm` Chlorine in hot condition according to given reaction `NaOH + Cl_(2) rarrNaCl + H_(2)O + O_(2)` is -A. 48B. 60 gmC. 24 gmD. 30 gm
Answer» Correct Answer - B `2NaOH + Cl_(2) rarr 2NaCl + H_(2)O + (1)/(2) O_(2)` `1.2` mole `0.6` mole `n_(NaOH "pure required") = 1.2` `W_(NaOH "pure") = 1.2 xx 40 = 48 gm` `W_("impure sample") xx (80)/(100) = W_(NaOH "pure")` `W_("sample") = 60 gm`

Discussion

97830.	200 gm of an Oleum sample labeled as 109 % is mixed with 518 gm water then the molality of final mixture is -A. 5.5 mB. 4 mC. 4.45 mD. 1 m
Answer» Correct Answer - C `(109) % = (100 + 9) % = 109 gm H_(2)SO_(4)` for each 100 gm Oleum From 200 gm Oleum `H_(2)SO_(4)` produced `= 109 + 109 = 218 gm` Water Remaining `= 518 = 500 gm` Molality `(m) = (218)/(98) xx (1000)/(500) = 4.45 m`

Discussion

97831.	18gm water is added in a 200 gm sample of Oleum labeled as `109%`. The new labelling of the final sample is :A. `106%`B. `103%`C. `100%`D. `0%`
Answer» Correct Answer - C Minimum labelling of Oleum `= 100%` All free `SO_(3)` is exhausted using `18gm` of water and pure `218gm H_(2)SO_(4)` formed.

Discussion

97832.	When 200 gm of an Oleum sample labeled as 109% is mixed with 300 gm of another Oleum sample labeled as 118%, the new labeling of resulting Oleum sample becomes -A. `112.6%`B. `114.4%`C. `113.5%`D. `127%`
Answer» Correct Answer - B After mixing 500 gm oleum sample consist of `320 gm SO_(3)` and `180 gm H_(2)SO_(4)` 100 gm oleum `rArr SO_(3) = (320)/(5) = 64 gm` `SO_(3) + H_(2)O rarr H_(2)SO_(4)` Moles `(64)/(80) " " (64)/(80)` Mass of `H_(2)O = (64)/(80) xx 18 = 14.4 gm` Labelling `= (100 + 14.4)%` `114.4 %`

Discussion

97833.	To 100 gm oleum sapmle Labelled as '118% oleum', 9 gm of water is added. Select the correct option/ options for final solutions.A. It contains only `H_(2)SO_(4)& SO_(3)`B. It contains `H_(2) SO_(4),SO& H_(2)O`C. New solutions will have labelling 100%D. % w/w of `SO_(3)` is neraly 36.7%
Answer» Correct Answer - A::D

Discussion

97834.	60 gr of oleum (labelled as 118%) is mixed with 11.8 gm of water. What will be that composition of final mixture?A. Only `H_(2)SO_(4)`, having mass 71.8 gmB. 118 gm of `H_(2)SO_(4)`C. 70.8 gm `H_(2)SO_(4)` & 1 gm waterD. 32 gm `SO_(3)` & 39.8 gm `H_(2)SO_(4)`
Answer» Correct Answer - 3 100 gm will contain 80 gm `SO_(3)` 60 gm will contain `80/100xx60=48 gm ` `{:(SO_(3),+,H_(2)O,to,H_(2)SO_(4)),(48 gm,,10.8 gm ,,58.8 gm),(0.6"mole",,"0.6 mole",,),(,,10.8" mole",,):}` water left =11.8-10.8=1 gm `H_(2)SO_(45)=58.8+12=70.8 gm `

Discussion

97835.	Arrange the following in the decreasing order of their basicity. Assign the reason :PH3, NH3, SbH3, AsH3, BiH3.
Answer» [Hint : NH₃ > PH₃> AsH₃ > SbH₃ > BiH₃]

Discussion

97836.	Determine the `pH` of an acidic buffer solution `(HA+NaA)` having acid and salt concentration respectively `0.1M` and `0.2M`. (Given `pK_(a) = 4.7, log 2= 0.30, log 5=0.70)`A. `4`B. `5`C. `3`D. None of these
Answer» Correct Answer - B Determine the `pH` …… `pH = 4.7+"log"(0.2)/(0.1)=5`

Discussion

97837.	Assign a possible reason for the following :(a) Stability of + 5 oxidation state decreased and that of + 3 oxidation state increases down the group 15 elements.(b) H2O is less acidic than H2S.(c) SF6 is inert while SF4 is highly reactive towards hydrolysis.(d) H3PO2 and H3PO3 act as reducing agents while H3PO4 does not.(e) Helium gas is used by scuba divers.
Answer» [Hint : (a) Due to inert pair effect. (b) Due to more bond dissociation enthalpy of O H as compared to S H bond. (c) Due to more steric hindrance offered by six F in SF₆ as compared to SF₄. (d) Due to presence of P H bond in them. (e) He is very less soluble in blood.]

Discussion

97838.	A buffer solution is made by mixing a weak acid HA `(K_(a) =10^(-6))` with its salt NaA in equal amounts. What should be amount of acid or salt that should be added to make 90 mL of buffer solution in which if 0.1 mole of strong acid are added into 1 L of this buffer solution then change in pH is unity ?A. 10 mmolesB. 22 mmolesC. 9 mmolesD. 11 mmoles
Answer» Correct Answer - D For buffer capacity of 0.1 we should have `((c+0.1)/(c-0.1))=10` where c is concentration of weak acid or salt in the buffer solution So, `c+0.1=10 c-1` so `9c=1.1` or `c=1.1/9` So, moles required for 90 ml solution =`1.1/9xx90xx10^(-3)` moles =11 m moles

Discussion

97839.	What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? a.`Kul(I)_(3)` b. `H_(2)ul(S)_(4)O_(6)` c. `ul(Fe)_(3)O_(4)` d. `ul(C )H_(3)ul(C )H_(2)OH` e. `ul(C )H_(3)ul(C )OOH`
Answer» In `KI_(3)`, since the oxidation numbers of `K` is `+1`, therefore, the average oxidation number of iodine `= -1//3`. But the oxidation number cannot be fractional. Therefore, we must consider its structure, `K^(+)[I-I larrI]^(Θ)`. Here a coordinate bond is formed between `I_(2)` molecule and `I^(Θ)` ion. The oxidation number of two iodine atoms forming the `I_(2)` molecule is zero while that of iodine forming the coordinate bond is `-1`. b. By conventional method: `overset(+1)(H_(2))overset(x)(S_(4))overset(-2)(O_(6)) or 2(+1)+4x+6(-2)=0` `x=+2.5` (wrong). But it is wrong beacuse all the four `S` atoms cannot be in the same oxidation state. By chemical bonding method, the structure of `H_(2)S_(4)O_(6)` is shown below: `H-Ooverset(+5)(-)underset(O)underset(\|\|)overset(O)overset(\|\|)(S)-overset(0)(S)-overset(0)(S)overset(+5)(-)underset(O)underset(\|\|)overset(O)overset(\|\|)(S)-OH` The oxidation number of each of the `S` atoms linked with each other in the middle is zero while that of each of the remaining two `S` atoms is `+5`. c. By conventional method. `overset(x)(Fe_(3))overset(-2)(O_(4))` or `3x+4(-2)=0` or `x=8//3` By stoichiometry `Fe_(3)O_(4) -= overset(+2)(Fe)overset(-2)(O).overset(+3)(Fe_(2))overset(-2)(O_(3))` Fe has oxidation number of `+2` and `+3`. d. By conventional method. `CH_(3)CH_(2)OH= overset(x)(C_(2))overset(+1)(H_(4))overset(-2)(O)` or `2x+6(+1)+1(-2)=0` or `x=-2` By chemical bonding `C_(2)` is attached to three `H` atoms (less electronegative than carbon) `Hoverset(2)(-)underset(H)underset(\|)overset(H)overset(\|)(C )overset(1)(-)underset(H)underset(\|)overset(H)overset(\|)(C )-OH` and one `CH_(2)OH` group (more electronegative than carbon). Therefore, oxidation number of `C_(3)=3(+1)-x+1(-1)=0` or `x=-2` `C_(1)` is, however, attached to one `OH` (oxidation number `-1)` and one `CH_(3)` (oxidation number `=+1)` of `C_(1)=+1+2(+1)+x+1(-1)=0` group, `:.` Oxidation number of `C_(1)=+1+2(+1)+x+1(-1)=0` `x=-2` e. By conventional method: `CH_(3)COOH=overset(x)(C_(2))overset(+1)(H_(4))overset(-2)(O_(2)) or 2x+4-4=0` or `x=0` By chemical bonding method: `C_(2)` is attached to three `H` atoms (less electronegative then carbon) `Hoverset(2)(-)underset(H)underset(\|)overset(H)overset(\|)(C )overset(1)(-)overset(O)overset(\|\|)(C )-OH` and one `-COOH` group (more electronegative tan carbon. Therefore, oxidation number of `C_(2)=3(+1)+x+1(-1)=0` or `x=-2` `C_(1)` is, however, attached to one oxygen atom by a double bond, one `OH` (oxidation number `=-1)` and one `CH_(3)` (oxidation number `=+1)` group. Therefore, oxidation number of `C_(1)=+1+x+1(-2)+1(-1)=0` or `x=+2`

Discussion

97840.	`1` mol of an ideal gas undergoes different thermodynamical process in `P-V` digram given below : If temperature at point `C` is `T K` then :A. Temperature at point `A` is `T K`.B. Temperature at point `A` is `2T K`.C. Temperature at point `B` is `T K`.D. Temperature at point `B` is `2T K`.
Answer» Correct Answer - B::D Process `CA` is isochoric `P_(A) = 2P_(C)` so `T_(A) = 2T` And process `AB` is isothermal `T_(A) = T_(B) = 2T`

Discussion

97841.	Consider the following redox reaction : `H_(2)O + AX + BY rarr HA + YO + BX_(2)` (unbalanced) It is also known that oxidation number of `X` is `-2` and neither `X` nor water is involved in the redox process. (In compound `BY, B` is cation and consider its oxidation number `lt 4`). Answer the following question based on the information given. Select the correct option(s) :A. The element `A` is under going reduction.B. The element `B` is under going reduction.C. The element `B` is under going oxidation.D. The element `Y` is under going oxidation.
Answer» Correct Answer - A::C::D

Discussion

97842.	Consider the following redox reaction : `H_(2)O + AX + BY rarr HA + YO + BX_(2)` (unbalanced) It is also known that oxidation number of `X` is `-2` and neither `X` nor water is involved in the redox process. (In compound `BY, B` is cation and consider its oxidation number `lt 4`). Answer the following question based on the information given. If the given reaction is balanced with smallest whole number coefficients, then:A. Sum of stoichmetric coefficient of all the components is `8`.B. Sum of stoichmetric coefficient of all the components is `9`.C. Stoichiometric coefficient of `AX ` is `6`.D. Stoichiometric coefficient of `HX ` is `2`.
Answer» Correct Answer - A::D `H_(2)O + 2AX + BY rarr 2HA + YO + BX^(2)`

Discussion

97843.	Assign oxidation number to the underlined elements in each of the following species: a.`NaH_(2)PO_(4)` b. `NaHul(S)O_(4)` c. `H_(4)ul(P_(2))O_(7)` d. `K_(2)ul(Mn)O_(4)` e. `ul(Ca)O_(2)` f. `Naul(B)H_(4)` g. `H_(2)ul(S_(2))O_(7)` h. `KAl(ul(S)O_(4))_(2).12H_(2)O`
Answer» a. Let the oxidation number of `P` be `x`. Writing the oxidation number of each atom above its symbol, we have `overset(+1)(Na)overset(+1)(H_(2))overset(x)(P)overset(-2)(O_(4))` Sum of oxidation numbers of varios atoms in `NaH_(2)PO_(4) =1(+1)+2(+1)+1(x)+4(-2)` `=x-5` But the sum of oxidation number of various atoms in `NaH_(2)PO_(4)` (netural) is zero `:. x-5=0` or `x=+5` Thus, the oxidation number of `P` in `NaHPO=+5`. b. `overset(+1)(Na)overset(+1)(H)overset(x)(S)overset(-2)(O_(4))` `:. 1(+1)+1(+1)+x+4(-2)=0` `or x=+6` Thus, the oxidation number of S in `NaHSO_(4)=+6`. c. `overset(+1)(H_(4))overset(x)(P_(2))overset(-2)(O_(7))` `:. 4(+1)+2(x)+7(-2)=0` or `x=+5` Thus, the oxidation number of `P` in `H_(4)P_(2)O_(7)=+5`. d. `overset(+1)(K_(2))overset(x)(Mn)overset(-2)(O_(4))` `:. 2(+1)+1(x)+4(-2)=0` or `x=+7` Thus, the oxidation number of `Mn` in `K_(2)MnO_(4)=+7`. e. Let the oxidation number of `Ca` be `x`. Since `O` in peroxides has an oxidation of `-1`. Thus, `overset(x)(Ca)overset(-1)(O_(2))` `:. x+2(-1)=0` or `x=+2` Thus, the oxidation number of calcium in `CaO_(2)=+2`. f. In `NaBH_(4), H` is present as hydride ion. Therefore, its oxidation number is `-1`. Thus, `overset(+1)(Na)overset(x)(B)overset(-1)(H_(4))` `:. 1(+1)+x+4(-1)=0` or `x=+3` Thus, the oxidation number of `B` in `NaBH_(4)=+3`. g. `overset(+1)(Na_(2))overset(x)(S_(2))overset(-2)(O_(7))` `:. 2(+1)+2(x)+7(-2)=0` or `x=+6` Thus, the oxidation number of `S` in `Na_(2)S_(2)O_(7)=+6`. h. `overset(+1)(K)overset(+3)(Al)(overset(x)(S)overset(-2)(O_(4)))_(2).12(overset(+1)(H_(2))overset(-2)(O))` or `+1+3+2x+8(-2)+12(2xx1-2)` or `x=+6` Alternatively, since `H_(2)O` is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in `H_(2)O` may be taken as zero. As such water molecules may be ignored while computing the oxidation number of `S`. `:. +1+3+2x-16=0` or `x=+6` Thus, the oxidation number of `S` in `KAl(SO_(4))2.12 H_(2)O=+6`.

Discussion

97844.	Two identical constaners A and B having same volume of an ideal gas at same temperature hace mass of the gas as `m_(1)` and `m_(2)` respectively and `2m_(1) = 3m_(2)`. The gas in each cylinder expands isomthermally to double of its voume. If change in pressure in A is `300 Pa`, then the change in pressure in B isA. `200 Pa`B. `300 Pa`C. `400 Pa`D. `500 Pa`
Answer» Correct Answer - A `2m_(1) = 3m_(2)` `P_(1)V = (m_(1))/(M) RT` and `P_(2)V = (m_(2))/(M)RT = (2)/(3)(m_(1))/(M)RT` `(P_(1))/(P_(2)) = (3)/(2)` The process is isothermal `(P_(1)V_(1) = P_(2)v_(2))` Hence, `DeltaP_(1) = P_(1) - (P_(1))/(2) rArr P_(1) = 2DeltaP_(1)` `DeltaP_(2) = P_(2) - (P_(2))/(2) = (P_(2))/(2) = (1)/(2)(2)/(30P_(1) = (P_(1))/(3)` `DeltaP_(2) = (2)/(3) DeltaP_(1) = (2)/(3).300` `DeltaP_(2) = 200 pa`

Discussion

97845.	Standard vaporization enthalpy of benzene at its boiling point is `30.8 kJ mol^(-1)`, for how long would a `100W` electric heater have to operate in order to vaporize a `100g` sample of benzene at its boiling temperature?
Answer» Standard vaporisation enthalpy of benzene `=30.8 kJ mol^(-1)` Power of electric heater `=100 W` Molar mass of `C_(6)H_(6)=72+6=78 am u` Heat required to vaporise `78 g C_(6)H_(6)=30.8xx1000J` heat required to vaporise `100 g C_(6)H_(6)` `=(30.8xx1000xx100)/78` `=39900 J` `100 W=100 J s^(-1)` `100 J` heat is given in `1 s` `39900` heat is given in `=(39900xx1)/(100)=399 s=6.6 min`

Discussion

97846.	Two wire of the same meta have same length, but their cross-sections are in the rati `3:1` . They are joined in series. The resistance of thicker wire is `10Omega` . The total resistance of the combination will beA. `20Omega `B. `30 Omega `C. `40 Omega `D. `50 Omega `
Answer» Correct Answer - C `(A_(1 ))/(A_(2)) =(3)/(1) implies R =( rho l)/( A) so (R _(1) )/(R_(2)) =(A _(2) )/(A_(1))=(1)/(3) ` `R_(1) =10 Omega ,R_(2) = 3R_(1)=30 Omega ` in series `R_(eq ) =R _(1) +R_(2) =10+30=40 Omega `

Discussion

97847.	In the circuit shown in fig. if both the bulbs `(B_1) and (B_2)` are identical A. their brightness will be the sameB. `B_(2)` will be brighter than `B_(1)`C. as frequency and that of `B_(2)` will becreaseD. Only `B_(2)` will glow because the capacitor has infinite impedance
Answer» Correct Answer - B Let `I_(1) and i_(2)` be current through `(B_1) and (B_2)` then `I_(1)xx sqrt(R^(2)xxX_(L)^(2))=220` `(I_2)/(I_1)=(sqrt(R^(2)xx X_(C)^(2)))/(sqrt(R^(2)xx X_(L)^(2)))=(sqrt(R^(2)+((1)/(c omega))^(2)))/(sqrt(R^(2)+L^(2)omega^(2)))` `(sqrt(R^(2)+((1)/(500xx10^(-6)xx2 pi xx 50))^(2)))/(sqrt(R^(2)+(10xx10^(-3)xx2 pi xx50)^(2)))` `=sqrt(R^(2)+40)//sqrt(R^(2)+9.87),I_(2)gtI_(1)` Bulb `(B_2)` will be brighter. As frequency and increases, `X_(C)` decrease `(X_L)` increases. `(I_2)` becomes less and `(I_1)` increases. `:.` brighter of `(B_1)` will increase and that of `(B_2)` decreases.

Discussion

97848.	A fly wheel of mass 12.5 kg and diameter 0.36 m rotating at 90 rpm has its speed increased to 720 rpm in 8s. Find the torque applied to flywheel.
Answer» ω₁ = 2πf₁ = 3π rad/sec ω₂ = 2πf₂ = 24 π rad/sec α = ω₂ – ω₁/ t = 8.243 rad/s² τ = 1α = (Mr² / 2) α τ = 1.669 Nm

97848.

A fly wheel of mass 12.5 kg and diameter 0.36 m rotating at 90 rpm has its speed increased to 720 rpm in 8s. Find the torque applied to flywheel.

Answer»

ω₁ = 2πf₁ = 3π rad/sec

ω₂ = 2πf₂ = 24 π rad/sec

α = ω₂ – ω₁/ t = 8.243 rad/s²

τ = 1α = (Mr² / 2) α

τ = 1.669 Nm

Discussion

97849.	A nucleus of mass m disintegrate into two pieces of mass \(\frac m3\) and \(\frac{2m}3\). The ratios of their de-Broglie wavelength will be -(1) \(\frac{\lambda_1}{\lambda_2} = \frac12\)(2) \(\frac{\lambda_1}{\lambda_2} = \frac21\)(3) \(\frac{\lambda_1}{\lambda_2} = \frac11\)(4) \(\frac{\lambda_1}{\lambda_2} = \frac43\)
Answer» Correct option is (3) \(\frac{\lambda_1}{\lambda_2} = \frac11\) Since their momentum will be same, so λ_db = same

97849.

A nucleus of mass m disintegrate into two pieces of mass \(\frac m3\) and \(\frac{2m}3\). The ratios of their de-Broglie wavelength will be -(1) \(\frac{\lambda_1}{\lambda_2} = \frac12\)(2) \(\frac{\lambda_1}{\lambda_2} = \frac21\)(3) \(\frac{\lambda_1}{\lambda_2} = \frac11\)(4) \(\frac{\lambda_1}{\lambda_2} = \frac43\)

Answer»

Correct option is (3) \(\frac{\lambda_1}{\lambda_2} = \frac11\)

Since their momentum will be same, so λ_db = same

Discussion

97850.	The de Broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron isA. hB. cC. 1D. 2
Answer» Correct Answer - B `E_"ph"/P_e =((hc)/lambda)/(h/lambda)`= c

Discussion

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