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limiting friction

The maximum friction that can be generated between two static surfaces in contact with each other. Once a force applied to the two surfaces exceeds the limiting friction, motion will occur. For two dry surfaces, the limiting friction is a product of the normal reaction force and the coefficient of limiting friction

1. More 2. Roughness 3. Resistance 4. Decreases 5. Less

Yeah , static friction is a self adjusting force because it comes into play when the body is lying over the surface of another body without any motion. (The body is at rest). ... It is known as self adjusting force because neither it has fixed magnitude nor direction.It adjusts according to the applied force.

option a is the correct answer of the given 6

please post clearly. Its very difficult to get your question.

. S(n th) = u +a/2(2n-1). S(5 th) = 0 + 8/2(2*5 - 1) = 4(10-1) = 4*9 = 36m

how a =10

please specify one question

The de Broglie wavelength varies inversly with the mass of a particle. So if we consider the mass of an electron and mass of a cricket ball we will see that cricket ball has the most significant mass than an electron. So we can observe the de Broglie wavelength of electron whereas the de Broglie wavelength of cricket ball is unobservable.

Newton’s second law of motion states that: the net external force applied on an object is directly proportional to the rate of change of linear momentum of the object.

Let the external force be F, initial velocity be u ,final velocity be v, the time interval be t.

=> p1(initial momentum ) = mu

p2(final momentum )= mv

Therefore, (rate of change of momentum)

Δp=(mv-mu)/t

According to newton’s IInd law , F α Δp

=>F =k * (mv-mu)/t { k is a constant }

In S.I.Units, the value of k is 1

So. F= m(v-u)/t

=> F = ma

Hence proved

Take equal volume of A and B, say V.Weight of A = Weight of V/2 volume of water.Similarly, weight of B = Weight of 3V/4 volume of water.As density is weight per unit volume, the required ratio will beD(A)/D(B)=(V/2)/(3V/4) =3/4Thus, density of A : density of B = 2:3

GravitationalForce– Weakestforce; but infinite range. ( Not partofstandard model)

Weak NuclearForce– Next weakest; but short range.

ElectromagneticForce– Stronger, with infinite range.

Strong NuclearForce– Strongest; but short range.

No two magnetic lines of force intersect each other.

The field lines emerge from North Pole and merge at the South Pole.

The direction of field lines is from its south pole to its north pole.

Magnetic field lines are the closed curves.

(1) P1=100 WV1 = 220 V

P = V×IP = V × (V/R)P = V^2/R

100 = (220×220)/R1R1 = 484 ohm

(2) P2 = 60 WV2= 220 V

60 = (220×220)/R2R2 = 806.6 ohm

Hence R2 is greater than R1

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We have given two lamps in such a way : Power of 1st lamp, P₁ = 40W , voltage of 1st lamp, V₁ = 220V power of 2nd lamp , P₂ = 60W , voltage of 2nd lamp ,V₂ = 220V

we know, one things , Power = V²/R [ when potential difference is same then consider P = V²/R ] R = V²/P so, resistance of 1st lamp , R₁ = V₁²/P₁ = (220)²/60 = 4840/6 = 2420/3Ωresistance of 2nd lamp , R₂ = V₂²/P₂ = (220)²/40 = 48400/40 = 1210Ω

energy consumed by lamps in one hour = Energy consumed by 1st lamps in one hour + energy consumed by 2nd lamps in one hour = P₁ × 1hour + P₂ × 1 hour =(P₁ + P₂) × 1 hour = (60W + 40W) × 1 hour = 100 Wh = 100/1000 KWh [ ∵ 1 KWh = 10²Wh ]= 0.1 KWh

It is a property by which an element can exist in different forms having different physical properties or we can say existence of two or more different physical forms of a chemical element.The different physical forms in which an element can exist are known as allotropes of that element.There are three allotropes for carbon:i)Fullerene ii)Diamond iii)Carbon

Power = P = 100 WVoltage = V = 220vResistance = R

P = V²/R100 = 220×220/RR = 220× 220/100 = 484Ω==========================Power = P = 60 WVoltage = V = 220vResistance = R

P = V²/R60 = 220× 220/RR = 220× 220/60 = 806.7ΩAs the resistors are connected in parallel ,total resistance = 1/R

1/R = 1/484 + 1/806.7 = 806.7 + 484/484×806.7 = 1290.7/390442.8R =390442.8/1290.7 = 302.5 ohms Total resistance= 302.5Ω

I = current

V = IR220 = I× 302.5

I = 220/302.5 = 0.73 A

The current drawn is 0.73 A

Resistance of 1st lamp, R1 = V1²/P1 = (220)²/100= 484 ohm

Resistance of 2nd lamp, R2 = V2²/P2 = (220)²/60= 806.6 ohm

Lamps are connected in parallel combination, Req = R1R2/(R1 + R2)= 484 × 806.6/(484 + 806.6)= 390394.4/1290.6= 302.5 ohm

so, I = V/Req = 220/(302.5)=.72 Amp

a.Joule's law of heating states that the heat produced in a wire is directly proportional to the product of sqaure of current flowing through it and the resistance offered by it.b. for the first bulb, P1=100W, V=220VI1V=100or, I=100/220= 0.45Afor 2nd bulb,P2=60W, V=220VI2=60or, I2=60/220=0.27A

Resistance of 1st lamp, R1 = V1²/P1 = (220)²/100= 484 ohm resistance of 2nd lamp, R2 = V2²/P2 = (220)²/200= 242 ohm.

if lamps are connected in series , Req = R1 + R2 = 484 + 242 = 726 ohmso, I = V/Req = 220/726 = 20/66 Amp = 10/33 = 0.3 amp

if lamps are connected in parallel combination, Req = R1R2/(R1 + R2)= 484 × 242/(484 + 242)= 484 × 242/726= 44 × 242/66 = 4 × 242/6 = 484/3 ohm

so, I = V/Req = 220/(484/3) = 660/484 = 60/44 = 15/11 = 1.36 Amp

in series it is rated 0.3 ampere and in parallel it is rated 1.36 ampere

When in seriescurrent will be same , Voltage will be different

Power = P = 100 WVoltage = V = 220vResistance = R

P = V²/R100 = 220×220/RR = 220× 220/100 = 484Ω

Power = P = 60 WVoltage = V = 220vResistance = R

P = V²/R60 = 220× 220/RR = 220× 220/60 = 806.7Ω

As the resistors are connected in parallel ,total resistance = 1/R

1/R = 1/484 + 1/806.7 = 806.7 + 484/484×806.7 = 1290.7/390442.8R =390442.8/1290.7 = 302.5 ohms Total resistance= 302.5Ω

I = current

V = IR220 = I× 302.5

I = 220/302.5 = 0.73 A

The current drawn is 0.73 A

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Power = P = 100 WVoltage = V = 220vResistance = R

P = V²/R100 = 220×220/RR = 220× 220/100 = 484Ω==========================Power = P = 60 WVoltage = V = 220vResistance = R

P = V²/R60 = 220× 220/RR = 220× 220/60 = 806.7Ω=========================As the resistors are connected in parallel ,total resistance = 1/R

1/R = 1/484 + 1/806.7 = 806.7 + 484/484×806.7 = 1290.7/390442.8R =390442.8/1290.7 = 302.5 ohms Total resistance= 302.5Ω

I = current

V = IR220 = I× 302.5

I = 220/302.5 = 0.73 A

The current drawn is 0.73 A

magnitude of i+j and i-j are equal having value= √1²+1² = √2

so, direction of both vectors are i+j/√2 and i-j/√2

now component of 2i+3j in the direction of i+j/√2 and i-j/√2 are ,

(2i+3j)•(i+j)/√2 = (2+3)/√2 = 5/√2 (2i+3j(i-j)/√2 = (2-3)/√2 = -1/√2

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Radius is = 2f or f is = R/2 so given radius is 32 cm and focal length will be 32/2 = 12cm hence lens is convex lens it's focal length is +ve

Radius of curvature R = 32 cmR =2ff=R/2f=32/2f=16 cm

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CONVEX LENS

Lens formula is 1/f = 1/v - 1/u

given that f=12 cm , u= -18 cm , v=?

Applying the formula we get:

1/12=1/v - (-1/18)=> 1/v=1/12+(-1/18)=>1/v=1/12 - 1/18=>1/v= 1/36so v = 36 cm

Thus the image is at a distance of 36 cm from the lens.The positive sign indicates it is on other side of lens as the object and so it is a virtual image.

For size.magnification formula is:m=h'/h=v/u (h'=image height,h=object height,v=image distance,u=object distance)h'/10= 36/-18h'= (36/-18) * 10h'= - 20cm

so,Position of image is 36 cm on other side of lens as object.Nature is virtual ImageSize of Image is 20 cm,which means it is magnified 2 times and is an erect image.

effective focal length 1/F = 1/f1 + 1/f2

=> 1/F = 1/15 + 1/10=> 1/F = (2+3)/30=> 1/F = 5/30=> 1/F = 1/6=> F = 6 cm

when we have half the tube is immersed in the mercury i.e 40 cm, the air pressure(P) inside the tube should be equal to the mercury pressure in the other half.

hence P = ρgh = ρg40

Once the tube is removed with upper end closed, the air pressure inside will remain same. Now since the other end is open, this air pressure + the pressure of mercury remaining = the atmospheric pressure at the other end.

=> P₀ = P + ρgh₁

= ρgh + ρgh₁

= ρg(40 + 20)

= 60ρg

Hence atmospheric pressure will be equal to pressure of 60cm of mercury column.

Thank you so much

conservation of angular momentumIw =constant for solid2/5 mR²w= constant wR²= constant 2πfR²= constant R²/T = constant so, T is directly proportional to square of radius Thus if radius is decreased to half then,T'= T/4which is 24/4time period will be 6 hours.