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By convention, for a flat lamina or a plane surface the area vector is a vector whose magnitude is the area of the surface and whose direction points in a direction perpendicular to the surface. The area vector of a closed surface is always zero.

Alfred Bernhard Nobel Swedish 21 October 1833 – 10 December 1896) was a Swedish chemist, engineer, inventor, businessman, and philanthropist.invented dynamite

(3) potential energy

Option(a) momentum

The dimension of Plank's contant can be obtained from any of the equations in which it appears, e.g. h = E/ν, λ = h/p, etc.

[h] = [ML^2T^−1] _____________(1)

Angular momentuml=rxp

[l] = [r] [p] = [L] [M LT^−1] = [ML^2 T^−1] _____________(2)

Compare Eqs. (1) and (2). See that both the Planck's constant and angular momentum have the same dimensions.

option c is the answer.

If an object is atequilibrium, then the forces are balanced. Balanced is the key word that is used to describeequilibriumsituations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects atequilibrium must havean acceleration of 0 m/s/s.

Given:Density of HCl=1.17g/cm³

Molarity=(density/molar mass)x1000 moles per liter

molar mass of HCl=[1+35.5]=36.5g/mol

Molarity=(1.17/36.5)x1000 =32.05M∴Molarity of liquid HCl with density equal to 1.17 g/cc is 32.05

It means prolongation of a sound; resonance.

Hey Arpita, here is ur answer :- Reverberation is the phenomenon of overlapping of sound caused by multiple reflections. Reverberation causes the overlapping of several reflected waves. If the time gap between the reflected waves is so short that these cannot be distinguished. So we hear multiple noisy sounds.

thank u very much.

u r welcome ☺️ Arpita Mallick. Hope that u find it useful..

To hear an echo clearly, the time interval between the original sound and the echo must be at least 0.1 s.

Since the speed of sound in air is 344 m/s, the distance travelled by sound in 0.I s = 344 m/s x 0.1 s = 34.4 m.

So to hear an echo clearly, the minimum distance of the reflecting surface should be half this distance that is 17.2 m.

Because the distance between us and the obstacle should be 17.2m for hearing the echo.In small rooms there is no 17.2m distance. So we cant hear any echo.Whereas in a big hall there is 17.2m distance. So we can hear many echo.

Given---

Distance covered by the man in 1 second. = 1 m.∴ Distance covered by the man in 60 min.= 1× 60× 60 = 3600 m.Now,Distance covered by the man is the 20 min. = 0 m.[Since boy is taking the rest]

Distance covered by the man in last 60 min. = 1 m× 60× 60= 3600 m.

∴ Total Distance Covered by the man = 3600 + 0 + 3600 = 7200 m.

Total Time taken by the man = 60 min. + 20 min. + 60 min.= 140 min.= 140× 60 seconds.= 8400 seconds.

Now,∵ Average Speed = Total Distance/Total Time = 7200/8400 = 72/84 = 6/7 m/s.

∴ Average Speed of the man during the entire Journey is 6/7 m/s.

For the Equilbrium ,Fe = FgKq^2/ r^2 = Gm1m2/r^2q = charger = distancem1 = mass of earthm2 = mass of moonq^2 = Gm1m2on substituting the valueq = 0.86 x10^14 C=8.6*10^13C

maximum velocity is the area under the curve

= 1/2*(11*10) = 55m/s

The relative speed between the two trains is :

30 - 10 = 20m/s

This is the initial relative speed.

When brakes are applied the train slows down.

Given the acceleration = 2 m/s²

u = 20 m/s

With time the train comes to a sudden stop making the velocity 0.

V² = U² - 2as

0 = 400 - 4s

400 = 4s

S = 400/4

S = 100m

given :-Radius , r = 105time t = 2min= 120 second

Circumference or distance of circle = 2πr⇒2×22/7×105⇒ 660 m∴ speed = distance / time⇒660 m/ 120⇒5·5m/s

c ) is the correct answer

Let the height of the tower be H...so, total time is t = √2H/g

Displacement in 1st three seconds is

h3 = 1/2*g*(3)² = 9g/2 =45m

also displacement in nth second is

hn = g/2*(2√(2H/g)-1)

but h3 = hn

so, 45 = 5*(2*(√(2H/g) -1)=> 9 = 2*√H/5 -1 => 10 = 2√H/5=> 5 = √H/5 => H/5 = 25

so, H = 25*5 = 125m

The relative speed between the two trains is :

30 - 10 = 20m/s

This is the initial relative speed.

When brakes are applied the train slows down.

Given the acceleration = 2 m/s²

u = 20 m/s

With time the train comes to a sudden stop making the velocity 0.

V² = U² - 2as

0 = 400 - 4s

400 = 4s

S = 100m

u = 5/18*72 = 20 m/secs = 25 mv = 0 (as car is made to stop)applying 3rd equation of motionv2= u2+2as0 = 400+50aa = -8 m/sec2Retardation = 8 m/sec2Retarding Force Required = Mass*Retardation = 8*500 = 4000 NHence 4000 N Of Retarding Force Is Required To Stop The Car Under The Given Circumstances

distance covered in last 2 seconds = area under the curve from time = 5 sec to 7sec = 1/2*(10*2) = 10m

and area under the curve from t = 1 to 7 is 1/2*(10*2) +10*2 +1/2*(10*2)= 10+20+10 = 40m

so, fraction covered in last 2 sec = 10/40 = 1/4

option B

which instruments measure the total during a given time

the value of gravitational acceleration varies between different bodies but for earth and any object near it's surface is 9.8 m/s^2.

t=5secv=0m/sec

therefore, a =v-u/ta=0-5/5a=-1m/sec2

or Retardation = 1m/sec2

Gravitational acceleration(symbolized g) is an expression used in physics to indicate the intensity of agravitationalfield. It is expressed in meters per second squared (m/s2). At the surface of the earth, 1 g is about 9.8 m/s2.

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At the poles it is9.832m/s²9.832m/s²and at the equator it is9.780m/s²9.780m/s².You can use this formulag(ß)=g−ш²(Rcosß)g(ß)=g−ш²(Rcosß)Whereg(ß)g(ß)is acceleration due to gravity at a latitudeßß,g=9.832m/s²g=9.832m/s²,шшis angular speed of the earth,RRis the radius of the earth.

distance covered in last 2 seconds = area under the curve from time = 5 sec to 7sec = 1/2*(10*2) = 10m

and area under the curve from t = 1 to 7 is 1/2*(10*2) +10*2 +1/2*(10*2)= 10+20+10 = 40m

so, fraction covered in last 2 sec = 10/40 = 1/4

option B

but in first 3 seconds it will have some velocity u which you have neglected

Let the total distance be D.

the point transversed half the distance by velocity 6m/s.

Then the time required by it to do so = (D/2)/6=D/12

The remaining half was transversed by travelling at 4m/sfor halftime and 6m/svelocity for another half time.

Let the time to cover the second half of the journey = t

Then4t/2+8t/2= D/2(Second Half of journey)

So t=D/(4+8)=D/12

The total time required for the journey is (D/2(6))+(D/(4+8)

The Average or mean Velocity = D/((D/2(6)) + (D/(4+8 ) =1/((1/2(6)) + (1/(4+8 ) ==2(6)(4+8)/(12+4+8)=12(12)/(24)=6m/s

(for your reference)Let the total distance be D.

Let the point transversed half the distance by velocity V1.

Then the time required by it to do so = (D/2)/V1=D/2V1

The remaining half was transversed by travelling at V2for halftime and V3velocity for another half time.

Let the time to cover the second half of the journey = t

ThenV2t/2+V3t/2= D/2(Second Half of journey)

So t=D/(V2+V3)

The total time required for the journey is (D/2V1)+(D/(V2+V3))

The Average or mean Velocity = D/((D/2V1) + (D/(V2+V3)) ) =1/((1/2V1) + (1/(V2+V3)) ) =2V1(V2+V3)/(2V1+V2+V3)----------(ans)

A free-falling object has an acceleration of9.8 m/s/s, downward (on Earth).

Flat surfaces reflect sounds that tend to reverberate, or echo. ALL of the surfaces in a concert hall are designed to control sounds, to push the sounds out, but not have them bounce around so much that the sound gets muddled and irritatingly difficult to hear properly.

(i)

Ceiling and walls are made curved so that sound after reflection reaches the target audience

The expressions or formulae which tell us how and which of the fundamental quantities are present in a physical quantity are known as theDimensional Formula of the Physical Quantity. Dimensional formulae also help in deriving units from one system to another. It has many real-life applications and is a basic aspect of units and measurements.

Suppose there is a physical quantity X which depends on base dimensions M (Mass), L (Length) and T (Time) with respective powers a, b and c, then its dimensional formula is represented as:

[MaLbTc]

A dimensional formula is always closed in a square bracket [ ]. Also, dimensional formulae of trigonometric, plane angle and solid angle are not defined as these quantities are dimensionless in nature