OSSAC TGT Science PCM Mock Test - 10 - OTET MCQ

Teachers are very closely linked to their students and motivate them to increase their knowledge and to develop confidence in them.

The magnetic field due to a circular loop is given by:
B= μ₀​​2πi​/4πr
=10⁻⁷×2π×2/0.0157 ​
=8×10⁻⁵ Wb/m²

F=-eE .
we know that,
F= ma ,
then
a=F/m.
a=-eE/m.

Here, I = 12.5 A, R = 3 cm = 3 x 10^-2 m 
x = 4 cm = 4 x 10^-2 m
∴ 

If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.Thus null point will move to fifth wire.

Light refracts as it passes through a material. It's direction changes, but it still passes through. Light travels at a maximum speed -- the speed of light in a vacuum, but when traveling in anything else it slows down. Different materials slow the speed of passing light at different rates. This property of matter is called the refractive index. When refracting, light doesn't change it's frequency, but since it changes it's speed, it must also change it's wavelength (it gets squished or elongated). Frequency, wavelength, and speed are all related, so if one property changes, another must as well.

A thick Plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face, and we have to determine the value of ' x'
According to the picture, the ray of light gets refracted at the interface between the air and the lens from the object 'p' and 'I' is the refracted image of 'p'
object distance 'u'= BO
Image distance 'v'/'x' = BI
we know
n2/v - n1/u = (n2- n1) / R
or 1/v - 1.5/(-3 )= (1-1.5)/ (-5) [where n2= 1, n1= 1.5 ,u= -3 ,R= -5]
so, 1/v = -6/15
or, v = -2.5
so, x is equal to 2.5 cm

The Critical Angle
a light ray is in the more dense medium and approaching the less dense medium.
The angle of incidence for the light ray is greater than the so-called critical angle.
When the angle of incidence in water reaches a certain critical value, the refracted ray lies along the boundary, having an angle of refraction of 90-degrees. This angle of incidence is known as the critical angle; it is the largest angle of incidence for which refraction can still occur. For any angle of incidence greater than the critical angle, light will undergo total internal reflection.

Iodine can gain only one electron.
I + e^- → I^-

Only a basic solution will change the colour of pH paper to greenish-blue.

and out of all these, only antacid is a base.

lemon juice and vinegar are acid

common salt is neutral 

Atomic mass of Y = (Atomic mass of X + Atomic mass of Z)/2

81 = (37 + Z) /2

162 -37 = Z

Z = 125

Air is a mixture of gases consisting of approximately 4/5 Nitrogen and 1/5 Oxygen.

An emulsion s a colloidal solution formed by mixing of two immiscible liquids.

In order for cells to function and survive, their organelles work together to carry out specific tasks and perform specific roles. Each organelle has its own role that contributes to the survival of the cell. This is called thr Division of labour. 

just apply the formula of (a+b+c)2 = a²+b²+c²+2ab+2bc+2ca

Required distance 

There are 366 days in a leap year, i.e, 1 more than a normal year.

Now, 52 weeks make up 344 days (52 x 7 = 344)
That means that we already have 52 sundays for sure.

Then, we are left with 2 days. Now, these days can be any from a pair of- mon-tues,tues-wed,wed-thurs,thurs-fri,fri-sat,sat-sun,sun-mon. Here favourable cases are sat-sun and sun-mon i.e, 2 cases and total number of cases is 7.

So, Probability=number of favourable cases/Total number of cases.

Therefore, Probability= 2/7.

y + sin y = 5x
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)

9 sec² A - 9 tan² A
= 9( sec² A - tan² A)
= 9 × 1
= 9

∵ AOB is a right triangle.
∴ Mid-point of AB is equidistant from A, O and B. 

AC=54cm, Radius=AO=54/2=27cm
Area of the bigger circle= πR²= 22/7 * 27 * 27=2291 (approx)
Diameter of the smaller circle=54-10=44
Radius=44/2=22
Area of the smaller circle= πr²=22/7 * 22 * 22=1521 (approx)
Area of the shaded region=Area of the bigger circle-Area of the smaller circle
= 2291-1521=770 cm²

|x| < 1 ⇒ -1 < x < 1
∴ No natural number exists between (-1, 1).

P(x,y) = [(e^t + e^-t)/2 , (e^t - e^-t)/2]
(e^t + e^-t)/2 = x --------------------------(1)
(e^t - e^-t)/2 = y --------------------------(2)
Adding (1) & (2)
2e^t = 2x + 2y
e^t = x + y
Eq (1) e^t + e^-t = 2x
e^t + 1/e^t = 2x
(e^t)² + 1 = 2x*e^t
(x+y)² + 1 = 2x(x+y)
x² + y² + 2xy + 1 = 2x² + 2xy
x² + y² + 1 = 2x²
(x²)/(1)² - (y²)/(1)² = 1  {which represents hyperbola equation} 

In a single toss, P(success) = 2/6 = 1/3 and P(non-success) = (1 − 1/3) = 2/3.
P(X=0) = P(non-success in the 1st draw and non-success in the second)
(2/3 × 2/3) = 4/9.
P(X=1) = P(success in the 1st toss and non-success in the 2nd) or (non-success in the 1st toss and success in the 2nd)]
(1/3 x 2/3) + (2/3 × 1/3) = 4/9.

 OE = x, OC = r, AO = h
∠AOC ≅ ∠QEC
AO / OC = QE/EC
(QE) = AO/OC(EC)
=h / r (r − x)
δx = 2πx (h / r(r − x))
=2 πhr (rx − x²)
δ'(x) = 0
x = r/2
derivating again
`delta''(r/2) = (2pih) / r(-2) < 0
r/2 is at maxima

Explanation:

Ratio of Complementary Angles:

- Complementary angles are two angles that add up to 90 degrees.
- The ratio of complementary angles is the ratio of the measures of the two angles that add up to 90 degrees.

Given:
Let the two complementary angles be x and y.

Therefore:
x + y = 90

Ratio of Complementary Angles:
The ratio of the two angles can be represented as:
x:y

Using the given information:
x + y = 90
x = 90 - y

Substitute x in the ratio:
(90 - y) : y

Simplify the ratio:
90 : y

Therefore:
The ratio of the two complementary angles is 90 : y.

Conclusion:
Since the ratio of complementary angles is 90 : y, the value of y is 1.

Answer:
C: 1