100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 237 InterviewSolution

11801.	Yesterday we didn’t have electricity, but we had some ______. We lit them to produce light. A) batteries B) candles C) papers D) fires
Answer» Correct option is B) candles

Discussion

11802.	I cut some bread with a ______. A) spoon B) fork C) knife D) plate
Answer» Correct option is C) knife

Discussion

11803.	Three pipes X, Y and Z can fill a pot in 6 hours. After working together for two hours, pipe Z is closed and pipe X and pipe Y fill the pot in 8 hours. In how much time the pot can be filled by pipe Z alone?1. 102. 123. 84. 9
Answer» Correct Answer - Option 2 : 12 Given: Three pipes X, Y and Z can fill a pot in 6 hours Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of the pot filled by (X + Y + Z) in 1 hours = \(\frac{1}{6}\) Part of the pot filled by these in 2 hours = \(\frac{2}{6}{\rm{}} = {\rm{}}\frac{1}{3}\) Remaining part = \(1 - \frac{1}{3}{\rm{}} = {\rm{}}\frac{2}{3}\) Time taken by X and Y in filling \(\frac{2}{3}\)rd part = 8 hours Time taken by X and Y in filling the whole pot = \(\frac{{8\; \times \;3}}{2}\) = 12 hours Part of pot filled by Z in an hour ⇒ \(\frac{1}{6} - \frac{1}{{12}} = \frac{1}{{12}}\) ∴ The required time is 12 hours.

11803.

Three pipes X, Y and Z can fill a pot in 6 hours. After working together for two hours, pipe Z is closed and pipe X and pipe Y fill the pot in 8 hours. In how much time the pot can be filled by pipe Z alone?1. 102. 123. 84. 9

Answer» Correct Answer - Option 2 : 12

Given:

Three pipes X, Y and Z can fill a pot in 6 hours

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of the pot filled by (X + Y + Z) in 1 hours = \(\frac{1}{6}\)

Part of the pot filled by these in 2 hours = \(\frac{2}{6}{\rm{}} = {\rm{}}\frac{1}{3}\)

Remaining part = \(1 - \frac{1}{3}{\rm{}} = {\rm{}}\frac{2}{3}\)

Time taken by X and Y in filling \(\frac{2}{3}\)rd part

= 8 hours

Time taken by X and Y in filling the whole pot

= \(\frac{{8\; \times \;3}}{2}\) = 12 hours

Part of pot filled by Z in an hour

⇒ \(\frac{1}{6} - \frac{1}{{12}} = \frac{1}{{12}}\)

∴ The required time is 12 hours.

Discussion

11804.	______ heats the room and consists of hollow metal container that fills up with hot water. A) radiator B) fire place C) bathtub D) sink
Answer» Correct option is A) radiator

Discussion

11805.	A tap can fill a container in 9 hours. Due to a leakage in its bottom, the container fills up in 10 hours. If the container is full, then in how much time the container will be emptied by the leakage?1. 70 hrs2. 80 hrs3. 90 hrs4. 100 hrs
Answer» Correct Answer - Option 3 : 90 hrs Given: A tap can fill a container in 9 hours With leakage container can be filled in 10 hours Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of the container emptied by the leak in 1 hours = (1/9) – (1/10) ⇒ (10 – 9)/90 ⇒ 1/90 ∴ The required time is 90 hours.

11805.

A tap can fill a container in 9 hours. Due to a leakage in its bottom, the container fills up in 10 hours. If the container is full, then in how much time the container will be emptied by the leakage?1. 70 hrs2. 80 hrs3. 90 hrs4. 100 hrs

Answer» Correct Answer - Option 3 : 90 hrs

Given:

A tap can fill a container in 9 hours

With leakage container can be filled in 10 hours

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of the container emptied by the leak in 1 hours = (1/9) – (1/10)

⇒ (10 – 9)/90

⇒ 1/90

∴ The required time is 90 hours.

Discussion

11806.	A leakage in the bottom of the container can empty the full container in 6 hours. An inlet pipe fills water at the rate of 4 L per minute. When the container is full, the inlet is open and due to the leakage the container is empty in eight hours. What is the capacity of the container?1. 5760 liters2. 96 liters3. 10 liters4. 24 liters
Answer» Correct Answer - Option 1 : 5760 liters Given: A leakage in the bottom of the container can empty the full container in 6 hours An inlet pipe fills water at the rate of 4 L per minute Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of container filled by inlet pipe in 1 hour = \(\frac{1}{6} - \frac{1}{8}{\rm{}} = {\rm{}}\frac{{4 \;- \;3}}{{24}}{\rm{}} = {\rm{}}\frac{1}{{24}}\) If there is no leak, the inlet pipe will fill the container in 24 hours. Capacity of the container = 24 × 60 × 4 = 5760 liters ∴ The capacity of the container is 5760 liters.

11806.

A leakage in the bottom of the container can empty the full container in 6 hours. An inlet pipe fills water at the rate of 4 L per minute. When the container is full, the inlet is open and due to the leakage the container is empty in eight hours. What is the capacity of the container?1. 5760 liters2. 96 liters3. 10 liters4. 24 liters

Answer» Correct Answer - Option 1 : 5760 liters

Given:

A leakage in the bottom of the container can empty the full container in 6 hours

An inlet pipe fills water at the rate of 4 L per minute

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of container filled by inlet pipe in 1 hour = \(\frac{1}{6} - \frac{1}{8}{\rm{}} = {\rm{}}\frac{{4 \;- \;3}}{{24}}{\rm{}} = {\rm{}}\frac{1}{{24}}\)

If there is no leak, the inlet pipe will fill the container in 24 hours.

Capacity of the container = 24 × 60 × 4 = 5760 liters

∴ The capacity of the container is 5760 liters.

Discussion

11807.	Consider following statement (A) Work done by pseudo force in non-inertial frame itself cannot be positive (B) Net work done by static friction on the system (consisting surfaces I contact) is always zero (C) Net work done by Kinetic friction on the system (consisting surfaces in contact) may be positive (D) Work done by kinetic friction on a body may be positive Select correct alternative :-A. A, B and CB. B and DC. A, B and DD. A, B, C and D
Answer» Correct Answer - A

Discussion

11808.	I jammed on the brakes and the stopped abmptly. (a) bike (b) car (c) cycle (d) bus
Answer» Correct answer is (c) cycle

Discussion

11809.	It is difficult to move a cycle along a road with its brakes on. Explain.
Answer» When the cycle is moved with its brakes on, wheels can only skid. There will be sliding friction. The sliding friction is more compared to rolling friction. Hence it is difficult to move a cycle with its brakes on.

Discussion

11810.	A block of mass `m = 2 kg` is connected to a a spring of force constant `k = 50 N//m`. Initially the block is at rest and tlhe spring has natural length. A constant force `F = 60 N` is applied horizontally towards, the maximum speed of the block (in `m//s`) will be (negelect friction).
Answer» Correct Answer - 6 For maximum speed, `(dv)/(dt) = 0 rArr a = 0` If at maximum speed, elongation is `x`, then `F = kx rArr x = (F)/(k)` Now, by `WET`. `-(1)/(2)kx^(2) + Fx = (1)/(2)mv^(2)` `V = (F)/(sqrt(mk)) = 6 m//s`

Discussion

11811.	The position of a paricle moving along `x-`axis depends on time as `x = 2(1 - e^(-31))`. Choose the correct statement(s).A. Total displacement of the particle is `2` unitsB. Total distance covered by particle is `2` units.C. Velocity of the particle will change its direction at some instant.D. Velocity of particle will not change its direction throughout the motion.
Answer» Correct Answer - A::B::D `x(0) = 0` `X(t rarr oo) = 2` `:.` Total displacement `= 2` units `v = (dx)/(dt) = 6e^(-3t)` ltbr `v gt 0 AA t` `:.` Displacement `=` Distance

Discussion

11812.	The coordinates of the point on the line joining the points (−3, 7, −13) and (− 6, 1, −10) which is nearest to the intersection of the planes 3x − y − 3z + 32 = 0 and 3x + 2y − 15z − 8 = 0 are (A) (9, 1, 7) (B) (7, 9, 1) (C) (−7, −1,− 9) (D) (−7, −9, −1)
Answer» Correct option is (C) (−7, −1,− 9)

Discussion

11813.	Shows the energy levels for an electron in a certain atom. Which transition shown represented the emission of a photon with the most energy? A. IIIB. IVC. Intensity of the characteristic S-rays depends on the electrical power given to the X-rays tubeD. II
Answer» Correct Answer - A `E = Rhc[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `E_((4 rarr 3))=Rhc[(1)/(3^(2))-(1)/(4^(2))]` `=Rhc[(7)/(9 xx16)] = 0.05 Rhc` `E_(4 rarr 2)=[(1)/(2^(2))-(1)/(4^(2))]` `=Rhc[(3)/(16)]=0.2 Rhc` `E_((2 rarr 1))=Rhc[(1)/((1)^(2))-(1)/((2)^(2))]` `= Rhc [(3)/(4)] =0.75 Rhc` `E_((1 rarr 3))=Rhc [(1)/((3)^(2))-(1)/((1)^(2))] = (8)/(9) Rhc = -0.9 Rhc` `:.` Thus, lll transition gives most energy.

Discussion

11814.	Find the synonymRespect A) esteem B) belief C) scorn D) contempt
Answer» Correct option is A) esteem

Discussion

11815.	Find the antonym of wordPut down A) patronize B) celebrate C) commend D) refuse
Answer» Correct option is C) commend

Discussion

11816.	Find the synonymPraise A) blame B) commend C) censure D) criticize
Answer» Correct option is B) commend

Discussion

11817.	The velocity of liquid (v) in steady flow at a location through cylindrical pipe is given by `v=v_(0)(1-(r^(2))/(R^(2)))`, where r is the radial distance of that location from the axis of the pipe and R is the inner radius of pipe. If R = 10 cm. volume rate of flow through the pipe is `pi//2 xx 10^(-2) m^3s^(-1)` and the coefficient of viscosity of the liquid is 0.75 N s`m^(-2)`, find the magnitude of the viscous force per unit area, in N`m^(-2)` at r = 4 cm.
Answer» Magnitude of viscous force, `F=eta A (dv)/(dt)` `rArr v=v_(0)(1-(r^(2))/(R^(2)))rArr (dv)/(dt)=-(2V_(0)r)/(R^(2))rArr sigma =eta.(2v_(0)r)/(R^(2))....(i)` Volume rate of flow, Q consider an annular element at r from axis, width dr. `dA=2pir dr, dQ=v.dA =v_(0)(1-(r^(2))/(R^(2)))2pirdr` `Q=intdQ=2piv_(0)[(r^(2))/2-(r^(4))/(4R^(2))]_(0)^(R)=(pi)/2R^(2) V_(0)rArr v_(0)=(2Q)/(piR^(2))` `:. (i) rArr sigma =eta(4Q)/(piR^(4))r, R=0.1 m` At r=0.04 m ,`sigma =(0.75) 4xx(pi)/2 xx10^(-2)xx0.04/(pixx10^(-4))=6 Nm^(-2)`

Discussion

11818.	Find the synonymStop A) cease B) eliminate C) commence D) commend
Answer» Correct option is A) cease

Discussion

11819.	A rectangular metal plate has dimensions of `10cmxx20cm`. A thin film of oil separates the plate from a fixed horizontal surface. The separation between the rectangular plate and the horizontal surface is `0.2mm`. An ideal string is attached to the plate and passes over an ideal pulley to a mass `m`. When `m=125gm`, the metal plate moves at constant speed of `5(cm)/(s)`, across the horizontal surface. Then the coefficient of viscosity of oil in `("dyne"-s)/(cm^2)` is (Use `g=1000(cm)/(s^2)`)
Answer» The coefficient of viscosity is the ratio of tangential stress on top surface of film (exerted by block) to that of velocity gradient (vertically downwards) of film. Since mass m moves with constant velocity, the string exerts a force equal to mg on plate towards right. hence oil shall exert tangential force mg on plate towards left. `:. eta=(F//A)/((v-0)//Deltax)=(125xx1000//10xx20)/((5-0)//0.2) =2.5 ` dyne-s/`cm^(2)`

Discussion

11820.	Assertion : Saheb was not allowed to play tennis in the neigh bourhood club. Reason : He had proper shoes but no racquet to play tennis. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true and R is not the correct explanation of A. c) A is true but R is false. d) A is false but R is true. e) Both A and R are false.
Answer» e) Both A and R are false.

Discussion

11821.	The raindrops are hitting the back of a man walking at a speed of `5km//hr`. If he now starts running in the same direction with a constant acceleration, the magnitude of the velocity of the rain with respect to him willA. gradually increaseB. gradually decreaseC. first decrease then increaseD. first increase
Answer» Correct Answer - B `omega^(2)R = R(d omega)/(dt), omega = (d theta)/(dt) = (omega_(0))/(1-omega_(0)t) implies T = (R )/(v_(0)) (1-e^(-pi))`

Discussion

11822.	Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = MV is imparted to the body at one of its ends what would be it angular velocity? A. V/LB. 2V/LC. V/3LD. V/4L
Answer» Correct Answer - C The impact force F = `(Deltap//Deltat)`, where `Deltap` = change of momentum of water of mass `Deltam` striking the ball with a speed v during time `Deltat`. Since water falls dead after collision with the ping-pong ball `" "Deltap=Deltamv implies F = v (Deltam)/(Deltat)` where `(Deltam)/(Deltat)` = rate of flow of water in the nozzle. Since the ball is in equilibrium `F - mg = 0 implies F = mg implies (vDeltam)/(Deltat) = mg implies (Deltam)/(Deltat) = (mg)/(v)`

Discussion

11823.	A solid sphere of mass `m` is lying at rest on a rough horizontal surface. The coefficient of friction between the ground and sphere is `mu`. The maximum value of `F`, so that the sphere will not slip, is equal to A. `7 mumg//5`B. `4 mumg//7`C. `5 mumg//7`D. `7 mumg//2`
Answer» Correct Answer - C The magnitude will decreases till the direction of the velocity with respect to man becomes vertical. It will increase thereafter.

Discussion

11824.	A particle is moving on a circle of radius `R` such that at every instant the tangential and radial accelerations are equal in magnitude. If the velocity of the particle be `v_(0)` at `t=0`, the time for the completion of the half of the first revolution will beA. `R//v_(0)`B. `(R//v_(0))(1-e^(-pi))`C. `(R//v_(0))e^(-pi)`D. `(R//v_(0))(1-e^(-2pi))`
Answer» Correct Answer - B Speed just before collision = `sqrt(2gl(1-costheta_(0)))` Speed just after collision = `e sqrt(2gl(1-cos theta_(0)))` From COE, `(1)/(2)me^(2)[2gl(1-costheta_(0))]=mgl(1-costheta)` `therefore theta = cos^(-1){1-e^(2)(1-costheta_(0))}`

Discussion

11825.	Solve:31/3 × 3-2/31. 1/√32. \(\sqrt[3]{3}\)3. \(\frac{1}{\sqrt[3]{3}}\)4. √3
Answer» Correct Answer - Option 3 : \(\frac{1}{\sqrt[3]{3}}\) Given: 31/3 × 3-2/3 Calculation: According to the question ⇒ 31/3 × 3-2/3 ⇒ 3^[^{1/3 + (-2/3)]} ⇒ 3^{(1/3 -}^2/3) ⇒ 3^(-1/3) ⇒ \(\frac{1}{\sqrt[3]{3}}\) ∴ Required value is \(\frac{1}{\sqrt[3]{3}}\)

11825.

Solve:31/3 × 3-2/31. 1/√32. \(\sqrt[3]{3}\)3. \(\frac{1}{\sqrt[3]{3}}\)4. √3

Answer» Correct Answer - Option 3 : \(\frac{1}{\sqrt[3]{3}}\)

Given:

31/3 × 3-2/3

Calculation:

According to the question

⇒ 31/3 × 3-2/3

⇒ 3^[^{1/3 + (-2/3)]}

⇒ 3^{(1/3 -}^2/3)

⇒ 3^(-1/3)

⇒ \(\frac{1}{\sqrt[3]{3}}\)

∴ Required value is \(\frac{1}{\sqrt[3]{3}}\)

Discussion

11826.	A 20 g bullet pierces through a plate of mass `M_(1)=1kg` and then comes to rest inside a second plate of mass `M_(2)=2.98` kg as shown in the figure. It is found that the two plates, initially at rest, now move with equal velocities, Find the percentage loss in the initial velocity of the bullet when it is between `M_(1)and M_(2)`. Neglect any loss of material of the plates due to the action of bullet A. `50%`B. `25%`C. `100%`D. `75%`
Answer» Correct Answer - B `"mu"+M_(1)xx0=M_(1)V+"mu"_(1)("for"I^(st)"plate")` `"mu"_(1)+M_(2)xx0=(m+M_(2))V("for"II^(nd)"plate")` `"from "I^(st)&II^(nd)"equation"u_(1)/u=3/4` `"so"%"loss"=(u_(1)-u)/uxx100%=25%`

Discussion

11827.	A mass of 2 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be:1. 250 π22. 25 π23. 200 π24. 100 π2
Answer» Correct Answer - Option 4 : 100 π2 Given: Mass (m) = 2 kg Radius (r) = 1 m v = 300 revolution/min Formula used: Angular speed (w) = 2πv Linear speed (v) = wr Kinetic energy = 1/2 mv² Calculation: v = 300/60 = 5 revolution/min Angular speed = 2π × 5 = 10π Linear speed (v) = wr = 10π × 1 = 10π Kinetic energy = 1/2 × 2 × 10π × 10π ⇒ 1/2 × 2 × 100 π² ⇒ 100 π² ∴ Required kinetic energy is 100 π²

11827.

A mass of 2 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be:1. 250 π22. 25 π23. 200 π24. 100 π2

Answer» Correct Answer - Option 4 : 100 π2

Given:

Mass (m) = 2 kg

Radius (r) = 1 m

v = 300 revolution/min

Formula used:

Angular speed (w) = 2πv

Linear speed (v) = wr

Kinetic energy = 1/2 mv²

Calculation:

v = 300/60 = 5 revolution/min

Angular speed = 2π × 5 = 10π

Linear speed (v) = wr = 10π × 1 = 10π

Kinetic energy = 1/2 × 2 × 10π × 10π

⇒ 1/2 × 2 × 100 π²

⇒ 100 π²

∴ Required kinetic energy is 100 π²

Discussion

11828.	Figure shows an At wood machine in which `m_(1)+m_(2)=M` is constant. Now a mass `Deltam` is transferred from `m_(2)` to `m_(1)`. Now a graph of `(a)/(g)` vs `Deltam` is plotted. Mark the most appropriate option (here `m_(1)gtm_(2)`) `:-` B. C. D.
Answer» Correct Answer - A Ans.(1) `a=((m_(1)-m_(2)))/(m_(1)+m_(2))grArr (a)/(g)=(m_(1)-m_(2))/(M)` Here `m_(2)darr` and `m_(1)uarr` so `(a)/(g) uarr`

Discussion

11829.	Two identical blocks `A` and `B` each of weight `20N` are placed on a frictionless horizontal plane. The portion of the string connected to `B` is vertical and that connected to `A` makes an angle of `53^(@)` with the plane . The minimum horizontal force `F` required to lift the blocks `B` is A. `12N`B. `16N`C. `32N`D. `36N`
Answer» Correct Answer - A Ans. (1) `Tcos53^(@)=F` & `T=20` `rArr20xx(3)/(5)=FrArrF=12N`

Discussion

11830.	Consider a quarter circular smooth rod AB fixed AB fixed in a vertical plane with centre of the circle at O. A bead of mass m slides without friction with the help of string starting from point A under the action of a constant force `F = 2 gm` as shown in figure. Work done by force F up till the moment when bead reaches at position B is : (Take `m = sqrt(2kg)` )A. `100 J`B. `200 J`C. `400 J`D. `50 J`
Answer» Correct Answer - C `W_(F) = (2 mg) (sqrt2R)`

Discussion

11831.	A bead of mass `m` slides without friction on a vertical hoop of radius `R`. The bead moves under the combined influence of gravity and a spring of spring constant `k` attached to the bottom of the hoop. For simplicity assume, the equilibrium length of the spring to be zero. The bead is released at the top of the hoop with negligible speed as shown . The bead, on passing the bottom point will have a velocity of `:` A. `2sqrt(gR)`B. `2sqrt(gR+(2kR^(2))/(m))`C. `2sqrt(gR+(kR^(2))/(m))`D. `sqrt(2gR+(kR^(2))/(m))`
Answer» Correct Answer - C From mechanical energy conservation `:` `mg(2R)+(1)/(2)k(2R)^(2)=(1)/(2)mv^(2)` `v=2sqrt(gR+(kR^(2))/(m))`

Discussion

11832.	A block of mass `1kg` slides down a curved track which forms one quadrant of a circle of radius `1m` as shown in figure. The speed of block at the bottom of track is `v=2 ms^(-1)`. The work done by the of friction is .
Answer» Correct Answer - B `mg1+w_(fk)=(1)/(2)(1)2^(2)-0` `w_(fk)=2-10`

Discussion

11833.	A block is projected up along the line of greatest slope of an inclined plane of angle of inclination `0(ltphi)` where `phi=` angle of friction. Which of the following friction time graph is CORRECT ?A. B. C. D.
Answer» Correct Answer - A First kinetic friction acts down the incline then static friction acts up the incline.

Discussion

11834.	Find the velocity (in m/s) with wich a block of mass 1 kg must be horizontally projected on a conveyer belt moving uniformly at a velocity of 3m/s, as shown, so that maximum heat is liberated. Coefficient of friction is 0.1.
Answer» `0^(2)=u^(2)=2mugxx8`

Discussion

11835.	A disc having radius `R` is rolling without slipping on a horizontal (`x-z`) plane. Centre of the disc has a velocity `v` and acceleration `a` as shown. Speed of point `P` having coordinates `(x,y)` isA. `(vsqrt(x^(2)+y^(2)))/R`B. `(vsqrt(x^(2)+(y+R)^(2)))/R`C. `(vsqrt(x^(2)-(y-R)^(2)))/R`D. none
Answer» Correct Answer - A `omega=v/R` Distance of P from origin. `r=sqrt(x^2+y^2)` Origin is instantaneous of rotation. So, `v_P=omegar=(vsqrt(x^2+y^2))/R`

Discussion

11836.	If y2 = 4ax find \(\frac{dy}{dx}\)
Answer» y² = 4ax Differentiate with respect to x. \(\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}\)

11836.

If y2 = 4ax find \(\frac{dy}{dx}\)

Answer»

y² = 4ax

Differentiate with respect to x.

\(\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}\)

Discussion

11837.	When brakes are applied to a moving car, the car travels a distance ‘s’ feet in ‘t’ secs is given by s = 20t – 40t2. When does the car stop?
Answer» Given s = 20t – 40t² v = \(\frac{ds}{dt}\) = 20 – 80t The car stops when the velocity = 0 ⇒ 20 – 80t = 0 ⇒ t = \(\frac{1}{4}\)sec

11837.

When brakes are applied to a moving car, the car travels a distance ‘s’ feet in ‘t’ secs is given by s = 20t – 40t2. When does the car stop?

Answer»

Given s = 20t – 40t²

v = \(\frac{ds}{dt}\) = 20 – 80t

The car stops when the velocity = 0

⇒ 20 – 80t = 0

⇒ t = \(\frac{1}{4}\)sec

Discussion

11838.	If y = cos (lnx). Show that x2y2 + xy1 + y = 0.
Answer» y = cos(lnx) \(y_1 = \frac{dy}{dx} = \frac{-sin(lnx)}{x}\) \(xy_1 = - sin(lnx)\) \(y _2 = \frac{dy}{dx^2} = \frac{- cos(lnx)}{x^2}+ \frac{sin(lnx)}{x^2}\) \(\therefore x^2y^2 = sin(lnx) - cos(lnx)\) Now, \(x^2y_2 + xy_1 + y = sin(lnx) - cos(lnx) - sin(lnx) + cos(lnx)\) = 0 Hence Proved.

11838.

If y = cos (lnx). Show that x2y2 + xy1 + y = 0.

Answer»

y = cos(lnx)

\(y_1 = \frac{dy}{dx} = \frac{-sin(lnx)}{x}\)

\(xy_1 = - sin(lnx)\)

\(y _2 = \frac{dy}{dx^2} = \frac{- cos(lnx)}{x^2}+ \frac{sin(lnx)}{x^2}\)

\(\therefore x^2y^2 = sin(lnx) - cos(lnx)\)

Now,

\(x^2y_2 + xy_1 + y = sin(lnx) - cos(lnx) - sin(lnx) + cos(lnx)\)

= 0

Hence Proved.

Discussion

11839.	If y = cos(x3) find \(\frac{dy}{dx}\)
Answer» Let y = Cos(x³) ∴ \(\frac{dy}{dx}\) = sin(x³). 3x²

11839.

If y = cos(x3) find \(\frac{dy}{dx}\)

Answer»

Let y = Cos(x³)

∴ \(\frac{dy}{dx}\) = sin(x³). 3x²

Discussion

11840.	If `y=acos(logx)+sin(logx),`prove that`(x^2d^2)/(dx^2)+ x(dy)/(dx)+y=0`
Answer» `y = acos(logx)+bsin(logx)` Let `logx = t =>1/xdx = dt=>dt/dx = 1/x` Then, `y = acost+bsint` `=>dy/dt = -asint+bcost` `:.dy/dx = dy/dt*dt/dx = (-asint+bcost)1/x` `=>xdy/dx = bcost-asint` Differentiating w.r.t. `x`, `=>x(d^2y)/dx^2+dy/dx = (-bsint-acost)dt/dx` `=>x(d^2y)/dx^2+dy/dx = -y/x` `=>x^2(d^2y)/dx^2+xdy/dx+y = 0`

Discussion

11841.	Find the point on the curve `y^2=4x`which is nearest to the point (2, 1).
Answer» `B(x_o,y_o)` normal at B passes through A `(y-y_o)=-1/(dy/dx)(x-x_o)` `(y-y_o)=-y_o/2(x-x_o)``1-y_o=-y_o +(x_oy_o)/2` `x_oy_o=2` `y_o^2=4x_o` `x_oy_oy_o^2=2*4x_o` `y_o^3=8` `y_o=2` `x_o=1` point(1,2).

Discussion

11842.	Two forces act at a point andare such that if the direction of one is reversed, the resultant is turnedthrough a right angle. Show that the two forces must be equal in magnitude.
Answer» Let `veca` and `vecb` are the two vectors.Then, their resultant force is `veca+vecb`. If `vecb` is reversed, then their resultant force will be `veca-vecb`. As the resultant turned through aright angle, their dot product should be `0`. `:. (veca+vecb)(veca-vecb) = 0` `=>vec.veca+veca.vecb-vecb.veca-vecb.vecb = 0` `=>vec.veca - veb.vecb = 0` `=>\|a\|^2 - \|b\|^2 = 0` `=> \|a\|^2 = \|b\|^2` `=> \|a\| = \|b\|` `:.` Magnitude of these two vectors are equal.

Discussion

11843.	Find the value of `lambda`, which makesthe vectors ` vec a , vec b`and ` vec c`coplanar, where ` vec a= hat i+3 hat j+4 hat k , vec b=2 hat i+lambda hat j+2 hat k`and ` vec c=4 hat i-7 hat j+10 hat k`
Answer» If `veca, vecb and vecc` are coplanar,`:. vecb(vecaxxvecc) = 0` `vecaxxvecc = \|[hati,hatj,hatk],[1,3,4],[4,-7,10]\|` `= (30-(-28))hati - (10-16)hatj + (-7-12)hatk` `=58hati+6hatj-19hatk` Now, `vecb.(vecaxxvecc) = 0` `(2hati+lambdahatj+2hatk)(58hati+6hatj-19hatk) = 0` `=>116+6lambda-38 = 0` `=>6lambda = -78` `=>lambda = -13`

Discussion

11844.	A tank can be filled up by two taps in 6 hours the smaller tap alone takes hours more than the bigger tap alone. find the the time required by each tap to fill the tank separately?
Answer» One tank can be filled up by two taps in 6 hours. The Smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required (in hours) by the smaller tap to fill the tank separately. Solution Let time taken to fill tank by small tap be x hours Time taken to fill tank by big tap be y hours Then, x−y=5 or x=5+y Smaller tap can fill in one hour = 1/x Bigger tap can fill in one hour = 1/y Then, 1/x+1/y = 1/6 x+y/xy = 1/6 xy=6(x+y) y(5+y)=6(5+y+y) 5y+y² =30+12y y² −7y−30=0 \(y=\frac{7\pm\sqrt{49+120}}{2}\) \(y=\frac{7\pm13}{2}\) y=10,−3 x=5+10=5+10=15 hours Hence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours \(Let time taken to fill tank by small tap be x hoursTime taken to fill tank by big tap be y hoursThen, x−y=5 or x=5+ySmaller tap can fill in one hour =x1Bigger tap can fill in one hour =y1Then,x1+y1=61xyx+y=61xy=6(x+y)y(5+y)=6(5+y+y)5y+y2=30+12yy2−7y−30=0y=27±49+120y=27±13y=10,−3x=5+10=5+10=15 hoursHence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours\)

11844.

A tank can be filled up by two taps in 6 hours the smaller tap alone takes hours more than the bigger tap alone. find the the time required by each tap to fill the tank separately?

Answer»

One tank can be filled up by two taps in 6 hours. The Smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required (in hours) by the smaller tap to fill the tank separately.

Solution

Let time taken to fill tank by small tap be x hours

Time taken to fill tank by big tap be y hours

Then, x−y=5 or x=5+y

Smaller tap can fill in one hour = 1/x

Bigger tap can fill in one hour = 1/y

Then, 1/x+1/y = 1/6

x+y/xy = 1/6

xy=6(x+y)

y(5+y)=6(5+y+y)

5y+y² =30+12y

y² −7y−30=0

\(y=\frac{7\pm\sqrt{49+120}}{2}\)

\(y=\frac{7\pm13}{2}\)

y=10,−3

x=5+10=5+10=15 hours

Hence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours

\(Let time taken to fill tank by small tap be x hoursTime taken to fill tank by big tap be y hoursThen, x−y=5 or x=5+ySmaller tap can fill in one hour =x1Bigger tap can fill in one hour =y1Then,x1+y1=61xyx+y=61xy=6(x+y)y(5+y)=6(5+y+y)5y+y2=30+12yy2−7y−30=0y=27±49+120y=27±13y=10,−3x=5+10=5+10=15 hoursHence, Small tap can fill the tank in 15 hours and bigger tap can fill in 10 hours\)

Discussion

11845.	In the figure, radius of the circle centered at O is 9 cm. OA = 15 cm. Semicircle with diameter O A cuts the circle with center O at D and BC is a tangent through B.1. What is the length of BC?2. If the line PD is perpendicular to OA, then what is the length of PD?
Answer» 1. BC² = OB × BA = 9 × 6 = 54 BC = √54 cm 2. OP × OA = r² \(OP=\frac{9^2}{15}=\frac{81}{15}\) PD² = OP x PA \(PD^2=\frac{81}{15}\times\frac{144}{15}\) \(PD=\frac{9\times12}{15}=\frac{36}{5}=7.2\) cm

11845.

In the figure, radius of the circle centered at O is 9 cm. OA = 15 cm. Semicircle with diameter O A cuts the circle with center O at D and BC is a tangent through B.1. What is the length of BC?2. If the line PD is perpendicular to OA, then what is the length of PD?

Answer»

1. BC² = OB × BA = 9 × 6 = 54

BC = √54 cm

2. OP × OA = r²

\(OP=\frac{9^2}{15}=\frac{81}{15}\)

PD² = OP x PA

\(PD^2=\frac{81}{15}\times\frac{144}{15}\)

\(PD=\frac{9\times12}{15}=\frac{36}{5}=7.2\) cm

Discussion

11846.	In the figure, AB is the tangent at B of the circle centered at O. How much is ∠OBA? How much is ∠AOB?
Answer» ∠OBA = 90° ∠AOB = 55°

11846.

In the figure, AB is the tangent at B of the circle centered at O. How much is ∠OBA? How much is ∠AOB?

Answer»

∠OBA = 90°

∠AOB = 55°

Discussion

11847.	Surfacetant molecules can cluster together as micelles, which are colloid sized cluster of molecules. Micelles from only above critical micelle concentration (CMC) and above centain temperature called K raft temperature. `DeltaH` of micelle formation can be positive or negative. Which is correct statement(s) about micelle formation?A. `DeltaS` of micelle formation is positiveB. the hydrophobic part lie towards interior of micelleC. the hydrophilic part lie towards surface of micelleD. `DeltaS` of micelle formation is negative
Answer» Correct Answer - BCD

Discussion

11848.	Give four uses of emulsion.
Answer» Various uses of emulsions are as follows `:` `a.` The concentration of sulphide ore by froth floatation process is based on emulsification. `b.` Cleansing action of soap and detergents is due to the formation of an emulsion between gas dirt and soap solution. ltbr. `c`. Milk is an emulsion of fat in water. ltbr. `d.` Cosmetics, lotions, creams, hair dyes, shampoos and many drugs, and ointments are emulsions. In the form of emulsions, these are more effective.

Discussion

11849.	What are micelles ? Give an example of a micelle system.
Answer» The particles of colloidal size formed due to aggregation of several units of stearate of soap molecules `(` surfactants `)` in a dispersion medium are called micelles. A concentrated solution of soap in water is a micelles system. Such substances are also called associated colloids. In other words. a micelles system `(` or associated colloid `)` behaves as a true solution in low concentration form and as a colloid in high concentration form. The micelles revert to individual ions on dilution.

Discussion

11850.	Comment on the statement that `"` colloid is not a substance but state of a substance `"`
Answer» The given statement is true. This is because the same substance may exist as a colloid under certain condition and as a crystalloid under certain other conditions. For example, `NaCl` in water behaves as a crystalloid, while in benzene, it behaves as a colloid `(` called associated colloid`)`. It is the size of particles which matters, `i.e.,` the state in which the substance exists. If the size of the particles lies in the range `1nm` to `1000nm`, it is in the colloid state.

Discussion

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