TS EAMCET Mock Test - 1 - JEE MCQ

(i) Effect due to rotation: Acceleration due to gravity is maximum at poles because there is no effect of earth's rotation at poles which are situated on the axis of rotation. Also, the earth is flattened at the poles due to which the distance from centre of the earth decreases and cause an increase in gravity. 

Acceleration due to gravity is minimum at the equator because there is maximum possible effect of the earth's rotation.

So, as we move from the equator to the poles, the acceleration due to gravity g increases. 

(ii) Effect due to radial distance from centre: The acceleration due to gravity is maximum at earth's surface. It decreases on moving towards the centre and above the surface.

Equation of straight line from t=0 to t=1 using straight line equation y=mx+c  where m is slope and c is the intercept
We get

When we convert any galvanometer into voltmeter,

V = I_g(R_g + R)

Let R_T be the effective resistance of R_g and R, connected in series.

V = I_g(R_T)

where, 

R_T = Total resistance of voltmeter

V = Range of voltmeter

I_g = Full scale deflection current
V ∝ RT

For a multi-range voltmeter, as we increase RT, range (V) would also increase.

So, the graph would be a straight line.

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

The magnetic field due to a circular coil at its centre of radius R and carrying current I is

The direction of magnetic field induction   at the centre is perpendicular to the plane of the circular coil and directed inwards.

The incident ray OA is refracted along BD by the lens. Singe the image coincides with the object. the refracted ray BD must retrace its path after reflection from the mirror, i.e., it falls normally on the mirror.
Let C be the point on the axis where BD produced meets the axis. It is clear that C is the centre of curvature of the mirror. Also is the imago of the object O formed by the lens alone. Thus
f = +10 cm and u = - 12 cm
The distance v of the image is obtained from the lens formula

which gives v = - 60 cm
Since x = 10 cm. R = v - x = 60 - 10 = 50 cm.

Hence, the focal length of the mirror = R/2 = 25cm

Total work done by the force F is the area under the force-displacement curve. This work is responsible to increase kinetic energy of the toy car.
Total work done by the force,

Hence, the kinetic energy of toy car at x = 11 m is,
Work done = Change in kinetic energy = 550 J
By energy conservation,
Potential energy gain by toy car at y_max = Kinetic energy at x = 11 m

For the block with mass m, the force experienced is the spring force.
∴ Spring force = Mass × acceleration = ma
For block with mass M, forces are spring force and the external F

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

we know,

Putting xe^x = t.
we get (xe^x + e^x) dx = dt
⇒ (x + 1) e^x.dx = dt

Clearly, (p→~p)∧(~p→q) is a contradiction.

Mixed doubles includes 2 men and 2 women.
Since 2 men are selected out of 6 men
⇒ number of ways = ⁶C₂
Also 2 women are selected out of 4 women
⇒ number of ways = ⁴C₂
But they also can be interchanged in 2 ways.
∴ Total no. of ways = ⁶C₂ x ⁴C₂ x 2

The given equation of curve is
y²+4x−6y+13=0
which can be written as
y²−6y+9+4x+4=0
⇒(y²−6y+9)=−4(x+1)
⇒(y−3)²=−4(x+1)
Put Y=y−3 and X=x+1
Then, we get
Y² = −4X
On comparing with the standard equation of parabola i.e., y²=4ax, we get
a=−1
Length of focus from vertex is a.
At focus X=a and Y=0
⇒x+1=−1
⇒x=−2
∴ y−3=0 ⇒ y=3
Hence, focus is (−2,3)

⇒ 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

= [10(21)]² − [5(11)]²
= (10 x 21)² − (5 x 11)²
= 44100 − 3025
= 41075 = An odd integer that is divisible by 5

= 
= 
= 

The equation of the circle touching the co-ordinate axes is
x² + y² - 2cx - 2cy + c² = 0
i.e. (x - c)² + (y - c)² = c² ...(1)
This touches the line

Perpendicular distance between centre (c, c) and line is

⇒ 7c - 12 = 5c or - 5c
⇒ 7c - 12 = 5c or 5c
⇒ c = 6 or c = 1
Hence, c = 1, 6

Given that probability that a teacher will give an unannounced test during any class meeting is 1/5.
The probability of not missing the test = 1 - 1/5 = 4/5
The probability of not missing both the tests = (4/5) × (4/5) = 16/25
Hence, probability that he will miss at least one test = 1 - 16/25 = 9/25

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

Let a and b be two points belonging to domain of f(x) such that
f(a) = f(b)

⇒ a = b or ab = 1 (Neglected as if  a ∈ N then b = 1/a does not belong to N)

Hence, a = b
⇒ The function is one-one.

For rectangular hyperbola a=b,
Hence (a−3)=−a,
So a=3/2