JEE TS EAMCET Mock Test - 1 Free Online Test 2026

Of the given options, option (1) satisfies the above condition.

The power delivered is the work done per second which again is equal to the total kinetic energy of the six bullets fired per second.

Hence power P=6 (kinetic energy of each bullet)

When both spheres are connected by a conducting wire then the charge q on smaller sphere will flow onto the larger sphere having charge Q.

Van de Graff generator work on this fact that charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.

Lens will form image of two sources at 70 cm from lens (i.e. screen)
From magnification of lens 

On solving, d = 0.3 cm
Now, fringes width β = λD/d

Two masses are moving with equal kinetic energy.

The ratio of linear momentum is

Since the current-carrying loop acts as an electromagnet, with application of Right-hand thumb rule, it can be stated that the face of the loop where the current flows in a clockwise direction behaves as a South pole and the other face where the current flows in an anti-clockwise direction behaves as a North pole.

Since, there is a force of attraction between the opposite poles, these coils will have a steady force of attraction between them.

When the block is displaced in one direction, the work done by the force is stored in both the springs as potential energy and when it is released, this potential energy is converted into kinetic energy.
Since at mean position, spring will be relaxed, so kinetic energy of block at mean position will be same as initial potential energy stored in springs.
So, the net loss in PE due to displacement of block from mean position = net gain in KE of the block at mean position.

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

In dienes, when the double bonds connected to each other by a single bond, this arrangement is known as conjugated.

Among the group 15 hydrides, the stability decreases on moving down the group, hence the order of stabilities of the hydrides of group 15 is

NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃

This is due to the increasing bond length down the group.
All the other statements are correct.

i.e. NCl₃ + 3H₂O → NH₃ + 3HOCl

Since the P-H bond is greater than the N-H bond, phosphine is a stronger reducing agent than ammonia.
NO is paramagnetic in the gaseous state. However, in the solid state, it dimerises, which results in the pairing of the odd electron.

we know,

⇒ Line is passing through (1,−1,3) and having direction ratios −3, −2, 1 i.e., 3, 2, −1

∴ Vector equation of the line is

Given equation can be rewritten as

To represent an ellipse, 

1. Start with
   sin⁻¹(1−x) − 2 sin⁻¹ x = π/2

2. Let θ = sin⁻¹ x,
   so x = sin θ and θ∈[−π/2, π/2].
   Then the equation becomes
   sin⁻¹(1−sin θ) − 2θ = π/2

3. Take sine of both sides:
   1 − sin θ = sin(π/2 + 2θ) = cos 2θ = 1 − 2 sin² θ

4. So
   1 − sin θ = 1 − 2 sin² θ
   ⇒ 2 sin² θ − sin θ = 0
   ⇒ sin θ (2 sin θ − 1) = 0
   ⇒ sin θ = 0 or sin θ = 1/2
   ⇒ x = 0 or x = 1/2

5. Check in the original:
   • x=0 ⇒ sin⁻¹(1) − 2 sin⁻¹(0) = π/2 − 0 = π/2 ✓
   • x=1/2 ⇒ sin⁻¹(1/2) − 2 sin⁻¹(1/2) = π/6 − 2·(π/6) = −π/6 ≠ π/2 ✗

Therefore the only solution is
 x = 0

For ellipse, equation: 
Let focus be (ae, 0) and directrix be x + ae = 0


And, for parabola: y² = 4aex

Length of latus rectum is 4ae.

According to the question,

Let h and r be the height and radius of cylinder.
Given that, dr/dt = 3 m/s, dh/dt = -4 m/s
Also, V = πr²h
On differentiating with respect to t, we get

At r = 4 m and h = 6 m
∴  = 80π cu m/s

Cases :

(i) If d=6 then seven digit numbers possible are = ⁵C₃⋅³C₃
[as a,b,c can be choosen from 1,2,3,4 or 5 & similarly e,f,g can be choosen from the unused 2 digit which are less than 6 & 0can be used]
(ii) If d=7 then numbers possible = ⁶C₃⋅⁴C₃
(iii) If d=8 then numbers possible = ⁷C₃⋅⁵C₃
(iv) If d=9 then numbers possible = ⁸C₃⋅⁶C₃
Add all cases, we get 1560

The equation of circle is x² + y² + 2gx + 2fy + c = 0 with centre (-g, -f) ......(1)
Comparing the equation x² + y² + 2x + 8y − 25 = 0 with (1), we get,
g = 1 and f = 4
∴ Centre of first circle C1 (−1, −4).

Comparing the equation x² + y² − 4x − 10y + 19 = 0 with (1), we get
g = −2 and f = −5
∴ Centre of second circle C₂ (2, 5).

 ⇒ It means circles intersecting each other.
Hence, number of common tangents = 2

Let f(x) = 2x³ - 3x² - 12x + 5, -2 ≤ ​​​​​​​​​​​​​​x ≤​​​​​​​ ​​​​​​​2
At a point of extremity, f'(x) = 0
Thus, 6x² - 6x - 12 = 0
Or, x² - x - 2 = 0
Or, (x + 1)(x - 2) = 0
Thus, x = -1 or x = 2
Now, f''(x) = 12x - 6
f''(-1) = 12(-1) - 6 = -12 - 6 = -18 < 0
As f''(-1) < 0, -1 is a point of maxima.
f''(2) > 0, 2 is a point of minima.
Thus, the maximum value occurs at x = -1.