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(a)we can see only that part of the moon which reflects light towards us

B ) is correct answer

yes above 2nd answer is right

let parts are sepereted at a distance of d

so repulsion force F= k q(Q-q)/d^2

here Q and d are constant so for maximum force differentiate the force w.r.t q

dF/dq = k/d^2[Q -2q]

for maximum force dF/dq=0

Q-2q=0

Q/q =2

Ans :- Material in thin plastic sheets that produces a high degree of plane polarization in light passing through it.

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thnx

At 90°, the intensity of the emergent light is zero.

thanks, from where did you got the answer

the charge q = - 5μC starts with a KE of 0.06 J from point C and stops at point d.

Initial total energy of the charge q = KE + PE = 0.06 + 9*10⁹ * 5 *10⁻⁶ * (-5 *10⁻⁶)/3 + 9*10⁹ * 5*10⁻⁶*(-5*10⁻⁶)/3 J = 0.06 - 0.075 - 0.075 J = - 0.09 J

Let ad = bd = x m

Final energy = PE (as KE = 0) = - 2 * 9*10⁹ * 5*10⁻⁶ * 5 * 10⁻⁶ /x J = - 0.450/x J

Hence, using Energy conservation we get: 0.450/x = 0.09 x = 5 mcd =√(x² - 3²) = 4 m

The change in kinetic energy when the ball reaches the bottom of the plane is 500J loss due to friction =100J (20%of 500J )net change in kinetic energy=400J 1/2×mass of ball×(velocity at base of the plane^2)-1/2×mass of the ball×(velocity at the top of the plane^2)=400J1/2×mass of the ball ×velocity at the base of the plane=400J (because velocity at the top of the plane is 0)1/2×5×(velocity at base ^2)=400Jvelocity at base =4 (root10)

Letthe fixed chargefor taxi= Rs x

Andvariable cost per Km= Rs y

Total cost= fixed charge + variable charge

Given that for a distance of 10 km, the charge paid is Rs 105

x + 10 y= 105… (1)

x= 105 – 10 y

Given that for a journey of 15 km, the charge paid is Rs 155

x + 15 y= 155

Plug the value of x we get

105 – 10 y + 15 y= 155

5y= 155 – 105

5y= 50

Divide by 5we get

y= 50 / 5= 10

Plug this value in equation (1) we get

x= 105 – 10 * 10

x= 5

People have to pay for traveling a distance of 25 km

= x + 25 y

= 5 + 25*10

= 5 + 250

=255Please like the solution 👍 ✔️

PCl5dissociates as:PCl5⇌ PCl3Cl2If p is the degree of dissociation at certain temperature under atmospheric pressure, thenInitial concentration:PCl5= 1PCl3= 0Cl2= 0At Equilibrium:PCl5= 1 – pPCl3= pCl2= pTotal number of moles at equilibrium = 1 – p + p + p = 1 + pPartial pressures of PCl5, PCl3and Cl2will be:p (PCl3) = p p / 1 + pp (Cl2) = p p / 1 + pp (PCl5) = (1 – p) p / 1 + pKp= p (PCl3) X p (Cl2) / p (PCl5)Kp=[(p p / 1 + p) X (p p / 1 + p)]/[(1 – p) p / (1 + p)]= p^2p / (1 – p)^2

Ans :- Fruits are the means by which angiosperms disseminate seeds. In common language usage, "fruit" normally means the fleshy seed-associated structures of a plant that are sweet or sour, and edible in the raw state, such as apples, bananas, grapes, lemons, oranges, and strawberries.

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Standard unitsareused in measurementbecause otherunitssuch as hand span are not appropriate because each person has different height.

because another units are used in measurement are appropriate from person to person . so standard unit are used in measurement

the mechanical equivalent of heat states that motion and heat are mutually interchangeable and that in every case, a given amount of work would generate the same amount of heat, provided the work done is totally converted to heat energy.

W=JH

J=W/H

W is the mechanical work which is equivalent to heat energy H. Now in SI unit both work and heat are measured in Joules. So mechanical equivalent of heat J= Joule/Joule = 1 (dimensionless)

In cgs units, heat is measured in calorie while work in erg. Now 1 calorie = 41800000 ergs approximately. So J= 41800000 ergs/calorie.

If 0.000001 m space is occupied by 1 bacterium then 1m will be occupied by 1/0.000001= 1000000 bacteria

No, The speed of light in vacuum is 2.998×10^8 m/s.

Speed of light is not possible to be more than the speed of light in vacuum. So, the minimum refractive index is 1.

V= 1.5×10 ^8m/s.

C= 3×10^8 m/s.

To find: n

solution: c/v.

n= 3×10^8 /1.5×10^8

n = 2.

Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.

ER. RAVI KUMAR ROY

Join the diagonal BD which divides quadrilateral ABCD in two triangles i.e ∆BCD & ∆ABD.

In ΔBCD,By applying Pythagoras Theorem

BD²=BC² +CD²

BD²= 12²+ 5²= 144+25

BD²= 169BD = √169= 13m

∆BCD is a right angled triangle.

Area of ΔBCD = 1/2 ×base× height

=1/2× 5 × 12= 30 m²

For ∆ABD,Let a= 9m, b= 8m, c=13m

Now,Semi perimeter of ΔABD,(s) = (a+b+c) /2

s=(8 + 9 + 13)/2 m= 30/2 m = 15 m

s = 15m

Using heron’s formula,Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 – 9) (15 – 9) (15 – 13)

= √15 × 6 × 7× 2 =√5×3×3×2×7×2=3×2√35= 6√35= 6× 5.92

[ √6= 5.92..]= 35.52m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30+ 35.5= 65.5 m²

Hence, area of the park is 65.5m²

Sound does not travel in vaccum because it needs a medium to travel.

This is a similar question-

Displacement of particle , y = 4sinπ(t/0.02 - x/75) here t = 0.1 sec and x = 50cm

so, y = 4sinπ(0.1/0.02 - 50/75)

y = 4sinπ(5 - 2/3)

y = 4sinπ(13/3) = 4sin(13π/3)

y = 4sin(4π + π/3) = 4sinπ/3 = 4 × √3/2

y = 2√3 = 2 × 1.732 = 3.464 cm

y = 4sinπ(t/0.02 - x/75)

differentiate with respect to t,

dy/dt = (4/0.02)πcosπ(t/0.02 - x/75)

v = 200πcosπ(t/0.02 - x/75)

at t = 0.1 sec and x = 50cm

v = 200πcosπ(0.1/0.02 - 50/75) = 200πcos(13π/3)

v = 200πcos(4π + π/3) = 200πcosπ/3

v = 200π × 1/2 = 100π = 314 cm/s or 3.14 m/s

1

2

Waves move energy from one place to another. In a progressive wave the wave front moves through the medium. There are two types of waves, transverse and longitudinal. Transverse waves are waves where the displacement of the particles in the medium is perpendicular to the direction the wave is travelling in.

A progressive wave is a wave where continuous energy transfer takes place between the crest and trough (transverse wave) or between rarefactions and compressions (long waves). These waves are associated with 2 trigonometric functions in terms of space and time

90km/hr

Solutions

An objectthat possessesmechanical energyis able todowork.

In fact,mechanical energyis often defined as the ability todowork.

Any objectthat possessesmechanical energy- whether it is in the form of potential energyor kineticenergy- is able todowork.

Momentumis the product of mass and velocity of a body, for a body mass never bezerobut velocitycanbezero ifa body is at rest, Soifa body is at restits momentum is zerobutif it isin rest at a height it possesses gravitational potentialenergyhencemechanical energy.

Momentumistheproduct of mass and velocity of a body, for a body mass never bezerobut velocitycanbezero ifa body is at rest, Soifa body is at restits momentum is zerobutif it isin rest at a height it possesses gravitational potentialenergyhencemechanical energy.

When all the three vectors are insameplane and forms a triangle and sum of two sides of the triangle is greater than the third side or the sum of any two vectors isequalto third one then all the three vectors arecollinearwill produce their resultant zero.

Motion is caused due to force. So an object moving at a uniform rate will keep moving till the force is applied on the object. Once the force is zero, there will be no motion and so the object will stop moving uniformly.

Yes,two vectorsof equal magnitude that are pointing in opposite directionswillsum tozero.Two vectorsof unequal magnitudecannever sum tozero. If they point along the same line, since their magnitudes are different, the sumwillnot bezero.

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