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Let the distance between the two stations is s meter. =>given v = 60 km/hr = 60 x 1000/3600 = 16.67 m/s By v = u + at =>16.67 = a x 120 =>a = 0.14 m/s^2 s = ut + 1/2at^2 =>s1 = 0 + 1/2 x 0.14 x (120)^2 =>s1 = 1000 m By v = u + at =>0 = 16.67 + a x 60 =>a = -0.28 =>By s = ut + 1/2at^2 =>s2 = 17.67 x 60 - 1/2 x 0.28 x (60)^2 =>s2 = 560.10 m Thus if s = s1+s1+ 1000 1st case:-(no interruption) =>T = t1+t2+ t3 =>T = 120 + 60 + 1000/16.67 =>T = 239.99 sec 2nd case:-(with interruption) =>T = t1+t2+t3 =>T = 120 + 60 + 1000/[(20 x 1000)/3600] =>T = 270 sec =>Delay = T(1st) - T(2nd) =>Delay = 270 - 239.99 =>Delay = 30.01 sec

f(x) = sinx +e^(-x)

f'(x) = cosx +(-e^(-x) = cosx-e^(-x)f"(x) = -sinx +e^(-x) .

now f'(0) = 1-1 = 0

f(0) = 0+1 = 1 f"(0) = 0+1 = 1. ; so f"(0) = f(0)

for maximum and minimum valuef'(x) = 0.=> cosx-e^(-x) = 0=> cosx = e^(-x)only solution is x = 0.

so, this is the point of minima.

hope it helps u

option (3) is correct.

0.16 mm is correct answer of this problem

Answer : RIn balance wheatstone bridge the galvanometer arm can be neglected so equivalent resistance is R

in series current will be same and in parallel connection voltage will be same

The voltage across A and B is same when they are connected in parallel and the current is same when they are connected in series....for series arrangement the net resistance is 10+5=15 ohm and for the parallel one it is (10×5)/(10+5)=50/15=3.33 ohm

thanks

when in parallel R=R1R2/R1+R2=5/3and in series R1+R2=125/3=R1R2/12R1R2=5/3*12=5*4=20R1=20/R2hence20/R2+R2=1220+R2^2=12R2R2^2-12R2+20=0R2=-b+-√b^2-4ac/2aR2=12+-√144-80/2R2=12+-8/2R2=4/2 or 20/2=2ohm and 10ohm

We know that -PTwo identical soap bubbles each of radius r and of the same surface tension t combine to form a new soap bubble of radius r.

The two bubbles contain air temperature. if the atmospheric pressure is pnot then find the surface tension t of the soap sol in terms of po,r,r.

Assume process is isothermal=2T/r2*50*10^-3/1*10^-3= 100N/m²

I

|velocity|≤|speed|

therefore, when speed is zero, then, velocity will be definitely zero. Hence, option (1) is correct.

Fe2O3 + Al -> Al2O3 + Fe + Heat This reaction is known as Thermit reaction which involves reduction of Ferric Oxide to Ferrum (Iron).

Rough surfaces have greater friction than smooth surfaces. Rough surfaces have grooves on them. When two rough surfaces are in contact, the grooves of one go inside the grooves of the other, thus making the movement difficult. Hence, roughness increases friction.

The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. Thefirst law of thermodynamicsstates that the change in internal energy of a system equals the net heat transferintothe system minus the net work donebythe system. In equation form, the first law of thermodynamics isΔU=Q−W.

Here ΔUis thechange in internal energyUof the system.Qis thenet heat transferred into the system—that is,Qis the sum of all heat transfer into and out of the system.Wis thenet work done by the system—that is,Wis the sum of all work done on or by the system. We use the following sign conventions: ifQis positive, then there is a net heat transfer into the system; ifWis positive, then there is net work done by the system. So positiveQadds energy to the system and positiveWtakes energy from the system. Thus ΔU=Q−W.

so angular velocity =3v/4a

all four charges are placed , each at a distance 'a ' from the origin.it means distance between two charges = √{a² + a² } = √2a

we also know, dipole moment is the system of two equal magnitude but opposite nature charges . so, we have to divide the charges as shown in figure for making dipole .

hence, there are three dipoles form. Let P = q(√2a) then, P₁ = P , P₂ = P and P₃ = 2P {as shown in figure.} now resolve the vectors P₁ ,P₂ and P₃ .

as shown in figure, vertical components of P₁ and P₂ is cancelled .and horizontal components of P₁ and P₂ is cancelled by horizontal component of P₃ .and rest part of dipole = vertical component of P₃ = 2Psin45° j ['j' shows direction of net dipole moment]hence, net dipole moment = 2q(√2a) × 1/√2 j = 2qa j

According to first law of thermodynamics,

Q = dU + W

Q= - 40J (heat released by gas)

W= - 20J (work done on gas)

Hence, - 40 = dU - 20

dU = - 20 J

Final internal energy - Initial internal energy = - 20

Therefore, Final Internal energy = 70 - 20= 50 J

(1) is correct option

Let the three equal mass fragments move along x, y, z direction with velocity vₒ (each have same k.e = E = ½mvₒ²) . conservation of momentum: before = after => 0 = mv̂ + m[vₒî + vₒĵ + vₒk̂] => v̂ = ivth fragment`s velocity = - [vₒî + vₒĵ + vₒk̂] ---------------------> (i) energy of explosion = k.e(f) - k.e(i) = k.e(f) - 0 = ½m(√3vₒ)² + 3(½mvₒ²) = 6½mvₒ² = 6E.

Masses of fragments: 1/5 * 1kg, 1/5* 1kg, 3/5 * 1kg ie., 0.2 kg, 0.2 kg, 0.6 kg

Linear momentum of theirs: 0.2 * 30m/s = 6 kg-m/s, 6 kg-m/s, 0.6 * v kg-m/s

The vector sum of the first two : 6√2 kg-m/s as they two linear momenta are perpendicular. This will be in North East direction, if the two pieces fly off in North and East directions respectively.

So linear momentum of the large piece is in the direction of South West direction as Linear momentum of all three pieces is 0 (as before explosion).

Velocity of the third piece = 6√2 kg-m/s / 0.6 kg = 10√2 m/s

The energy released is the K.E final..

1.5m/s is correct answer

1.5 is the best answer from my side

4)

m1v1 + m2v2 = mv5*0 + 35*v2 = 40*72v2 = 40*72/35 = 82.285 km/h.

is c option correct pls inform me

mass of 1 part = 3 kgby applying formulakE = p²/2m = 36

Masses of fragments: 1/5 * 1kg, 1/5* 1kg, 3/5 * 1kg ie., 0.2 kg, 0.2 kg, 0.6 kg

Linear momentum of theirs: 0.2 * 30m/s = 6 kg-m/s, 6 kg-m/s, 0.6 * v kg-m/s

The vector sum of the first two : 6√2 kg-m/s as they two linear momenta are perpendicular. This will be in North East direction, if the two pieces fly off in North and East directions respectively.

So linear momentum of the large piece is in the direction of South West direction as Linear momentum of all three pieces is 0 (as before explosion).

Velocity of the third piece = 6√2 kg-m/s / 0.6 kg = 10√2 m/s in South West dir.

mass of the bomb ,M=9kg

initial value of the bomb=0m/s

mass of smaller fragmet =3kg=m1 (let)

velocity of the smaller part just after explosion=16m/s

mass of the bigger fragmet =6kg=m2 (let)

let, velocity of the bigger part just after explosion= v

as there is no external force is acting on the bomb intially

so the momentum is conserved in all direction

i.e P(just after explosion)=P(just before explosion )

m1*16 + m2*v = M*0

=> v = -8m/s

so the kinetic energy of the fragment of mass 6kg = (1/2)*m2*v*v Joule

=(1/2)*6*64

=192 Joule

Let the particle be displaced from position A to B under the action of the given force thenAB=B-A (taking(0,0,0) as origin) =2i+3j-2kNow,work done on the particle =(force). (displacement) =(3i+j).(2i+3j-2k)=9

Lightweighting of cars, trucks, airplanes, boats, and space craft could lead to significant fuel savings.

Nano-bioengineering of enzymes is aiming to enable conversion of cellulose from wood chips, corn stalks, unfertilized perennial grasses, etc., into ethanol for fuel.

Nanoparticles are used increasingly in catalysis to boost chemical reactions.

Nano-engineered materials make superior household products such as environmental sensors, air purifiers, and filters; antibacterial cleansers; and specialized paints and sealing products, such a self-cleaning house paints that resist dirt and marks.