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```A peep into the toughest questions of BITSAT with solutions!

Table of Content

Mathematics

Physics

Chemistry

Logical Reasoning

English Proficiency

Useful Resources

BITS Pilani entrance exam commonly known as BITSAT, is one of the most sought-after engineering exams to gain admission in premier institute of BITS Pilani. The exclusivity of the exam lies in the fact that it is conducted in online mode and tests the entire personality of applicants. Unlike other top-notch entrances like the JEE or VITEEE, BITS entrance exam also examines the English proficiency and Aptitude of candidates.

BITSAT is generally based on the NCERT syllabus but demands clarity of concepts. The questions asked in the exam are not very tricky but are based on application of theoretical concepts.  Let us have a glance at some of the challenging questions of each section one by one:

Mathematics

The best way to excel in mathematics section is by having a very good speed. Moreover, the method of elimination is the best weapon. Instead of attempting to solve the question completely, one must try to eliminate the options by applying certain tricks. Some of arduous problems of mathematics are listed below:

Illustration 1: If 2 < x < 3, then

(a) |x-3| < |x-2|                                           (b) (x-3) > (x-2)

(c) (x-3)(x-2) < 0                                             (d) (x-3)/(x-2) > 0

Solution: If a, b ? R and a < b, then a < x < b.

This is true iff (x-a)(x-b) < 0

Alternatively, 2 < x < 3,

This implies (x-2) > 0 and (x-3) < 0

Hence, this clearly shows that (x-2) and (x-3) are of opposite signs.

Illustration 2: If If R = (√2 + 1)2n+1 and f = R - [R], then [R] equals

(a) f +1/f                                                        (b) f - 1/f

(c) 1/f - f                                                        (d) None of these

Solution: Let G = (√2 - 1)2n+1

Then R – G = (√2 + 1)2n+1 - (√2 - 1)2n+1 = 2 [2n+1 C1 (√2)2n + ….  ]

which is obviously an integer

Hence, [R] + f – G is an even integer

This gives f - G = 0

Hence f = G. Therefore, [R] = 1/f – f.

Physics

Physics is believed to be toughest section in BITSAT. Thus it is crucial to have a thorough understanding of topics. Students must not blindly follow some standard method to solve a problem, but should try to identify why a particular formula is applied in a specific case. Certain demanding problems of Physics are listed below:

Illustration 1: The equation of sound wave is

y = 0.0015 sin (62.4x + 316t).

Then the wavelength of this wave is

(a) 0.2 unit                                                                 (b) 0.1 unit

(c) 0.3 unit                                                                  (d) 0.4 unit

Solution: The given equation of sound wave is

y = 0.0015 sin (62.4x + 316t).

We know that the standard equation of wave motion is

y = A sin 2π (x/λ + t/T)

Hence, just comparing the given equation with the standard equation we get

2π/λ = 62.4

Therefore, the wavelength of the wave is λ = 2π / 62.4 = 0.1 unit.

Illustration 2: What is the magnitude of the magnetic force per unit length of a wire carrying a current of 5A and making an angle of 30° with direction of uniform magnetic field of 0.1T?

Solution: It is given in the question that the current I = 5A.

The angle θ = 30° and the magnetic field B = 0.1T.

Thus, the magnitude of magnetic force per unit length on a wire is given by

= IB sin θ

Substituting the values in the above formula we get,

= 5 x 0.1 x sin 30°

= 0.5 x 0.5

= 0.25 N-m.

Chemistry:

It is important to memorize al the chemical reactions and formulae in chemistry. The questions asked in the exam are not very tricky but are usually based on the application of formulae and chemical reactions. some of the sample problems are listed below:

Illustration 1: The volume of air needed for complete combustion of 1kg carbon at STP is

(a) 3333.35 L                                               (b) 6666.66 L

(c) 9333.35 L                                               (d) 9999.99 L

Solution: We know that C + O2 → CO2

Also, O2 required by 12 g of carbon = 22.4 L

Hence, O2 required by 1000 g of carbon = 22.4 x 1000/12 = 1866.67 L

Also, we know that oxygen is 1/5 th part of air

Hence, the volume of air = 5 x volume of O2 = 5 x 1866.67 L = 9333.35 L

Illustration 2: On adding 0.750 g of compound in 25 g of solvent, it lowered the freezing point of the solvent by 0.502°C. the molecular weight of the substance is [ the molecular depression of constant = 50.2°C per 100 g of solvent.

(a) 100                                                         (b) 200

(c) 300                                                         (d) 400

Solution: We know that the formula for molecular weight,

m = 100 x Kf x w / ?Tf x W,

where Kf = molecular depression constant = 50.2°C per 100g

w = weight of compound = 0.750g

?Tf = lowering of freezing point of solvent

= 0.502°C and

W = weight of solvent = 25g

Hence, m = (100 x 50.2 x 0.750) / (0.502 x 25) = 300.

Logical Reasoning:

Students often tend to ignore this section and focus on the three main subjects. This is not the correct approach. This section can fetch you good scores with little efforts. There are some fixed patterns of questions asked from this section. Hence, it is important to concentrate on this segment as well.

Illustration 1: A train 50 metres long passes a platform 100 metres long in 10 second. What is the speed of the train per second?

Solution: The train is given to be 50 metres long, hence l = 50m

Length of platform d = 100m and time t = 10sec

The total distance covered by the train to pass the platform is given by

s = l + d = 50 + 100 = 150 m

Hence, the speed of train is v = s/t = 150/10 = 15 m/s

Illustration 2: Insert the missing number in the following sequence

1, 3, 6, 10, 15, ….. 28, 36, 45

Solution: Just notice the given sequence

The difference of each successive and previous number has the sequencial increase of one as

3-1 = 2, 6-3 = 3, 10-6 = 4……

Hence, the sequence of differences is (2, 3, 4, 5, 6, 7, 8, and 9)

The next number after 15 in the series has to be 21. (Since 21-15 = 6)

English Proficiency:

This can be called as the most scoring section of BITSAT. You just need to have a strong base to be able to crack it. Questions asked in the exam include simple comprehension questions, synonyms and antonyms. Some sample questions are as follows:

Illustration 1: Select the pair of words which are related in the same way as the capitalized words are related to each other

MEANS: METHOD :: ?

(a) elegance : style

(b) air : kite

(c) teacher : pupil

(d) smoke :cigar

Solution: Elegance is related to style in the same way as means are related to methods.

Illustration 2: Choose the most appropriate synonym of the word

GROUSE

(a) brave                                                        (b) complain

(c) lazy                                                          (d) tired

Solution: Complain is the synonym of grouse.

It is not very difficult to crack BITSAT. Just a good strategy in the right direction can fetch you excellent scores!

Useful Resources

Bitsat Cutoff | Bitsat Rank Predictor | BITSAT Revision Notes | Study Material for BITSAT Exam Preparation

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