JEE TS EAMCET Mock Test - 9 Free Online Test 2026

The ray falls normally on AB is refracted undeviated into the prism and incident on face AC at an angle of incidence i=θ.

So, for total reflection at AC face,

Let frequency of second fork =x
We know that beat frequency =|f₁ − f₂|

On loading the fork with wax, its frequency decreases. So beat frequency will increase if,x=335 Hz and decrease if, x=347 Hz. According to the given condition, beat frequency decreases, so x=347 Hz

Charged particle can be considered at centre of a cube of side a, and given surface represents its one face. So, flux 

Let, v be the velocity of a wave.
Length of open pipe A be l₁.
 Length of closed pipe B be l₂.

Formula used:

nth overtone of an open pipe f_n = (n+1)f₀, where for open pipe fundamental frequency f₀ = v/2l
nth overtone of a closed pipe f_n = (2n+1)f₀, where for closed pipe fundamental frequency f₀ = v/4l.

Now, solving the problem

Since, it is given that frequencies are same,

Now, the ratio of fundamental frequencies,

For A, T = f = 2mg

2mg − mg = ma₁

∴ a₁ = g

For B,

From the force diagram shown in the figure,

2mg − mg = 3ma₂

a² = g/3

For C,

∴ 3mg − mg = 2ma₃

⇒ a₃ = g

So, a₁ = a₃ > a₂

Average force = Change in momentum/Time interval

Expansion work done by the gas  cal
 cal

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Sorption is the common term used for both absorption and adsorption. Absorption is the incorporation of a substance in one state into another of a different state (e.g., liquids being absorbed by a solid or gases being absorbed by water). Adsorption is the physical adherence or bonding of ions and molecules onto the surface of another molecule. The term 'sorption' is preferred when it is suspected that both processes are operative.

• The ozone layer is present in the stratosphere and the Taj Mahal is affected by acid rain.
• The second and fourth statement is incorrect because the Ozone layer is present in the lower portion of the stratosphere. Also, the Taj Mahal is affected by the presence of oxygen, chlorine, sulphur dioxide, and hydrogen. 

we have

y = e^cx
dy/dx = c. e^cx
y' = cy

We have,

a²+2b=7   …(i)

b²+4c=−7   …(ii)

c²+6a=−14   …(iii)

Adding equations (i), (ii) and (iii), we have

It is possible only when a=−3, b=−1 and c=−2.

So,

On differentiating with respect to x

On substituting the values in equation (i)

Let

Feasible solutions are

Clearly, the solution region is the quadrilateral.

From P(4, 3) two tangents PT and PT′ are drawn to the circle x²+y²=9 with O(0, 0) as centre and r=3. To find the area of  ΔPTT′ .

Let R be the point of intersection of OP and TT′,

Then, we can prove by simple geometry that OP is perpendicular bisector of TT′.

The equation of chord of contact from a point (x₁, y₁) to a circle x²+y²=r² is xx₁+yy₁=r₂

Hence, the equation of chord of contact TT′ from the point (4, 3) to the circle x²+y²=9 is 4x+3y=9

The length of perpendicular from a point (x₁, y₁) on a line 

Now, OR= length of the perpendicular from O to TT′ is 

Now, radius of circle OT=3
In triangle OTR, by Pythagoras theorem, we have

The distance between the points (x₁, y₁) and (x₂, y₂) is 

∴ Area of the triangle PTT′ will be twice area of ΔPRT

The given equation reduces to (−ω)^m    = (−ω²)^m 

⇒ ω^m  = 1 

⇒ m = 3

N₁ + N₂ + N₃ + N₄ + N₅  =  Total ways – {when no odd}

Total ways = ⁹C₅

Number of ways when no odd number is selected is 0 (∵ only available even are 2,4,6,8)