NCERT Solutions for Exercise 15.1 Class 10 Maths Chapter 15 - Probability

NCERT Solutions for Exercise 15.1 Class 10 Maths Chapter 15 - Probability

Edited By Sumit Saini | Updated on Jul 11, 2022 07:00 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Maths exercise 15.1 - The experimental probabilities are based on the results of actual experiments as well as adequate recordings of the events. Furthermore, these probabilities are only 'estimations.' If we repeat the experiment 1000 times, we might get different results with different probability estimates.

The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as

NCERT solutions Class 10 Maths exercise 15.1- An elementary event is one that has only one outcome of the experiment.

The sum of the probabilities of all the elementary events in an experiment is one. An Impossible event is one that has no chance of occurring, i.e., P(E) = 0. The probability of an event ranges from 0 to 1, inclusive of 0 and 1. i.e.,

Complementary events are the only two possible outcomes of a single event. This is analogous to tossing a coin and seeing if it lands on heads or tails.

where E is representing event and is representing not E or complementary of the event E.

Along with Class 10 Maths chapter 15 exercise, 15.1 the following exercise is also present in the NCERT book.

Probability Class 10 Chapter 15 Exercise: 15.1

Probability of an event E + Probability of the event ‘not E’ = 1

$\\ E \cup E' = S \\ P(E \cup E') = P(S) = 1$

The probability of an event that cannot happen is 0. Such an event is called an impossible event.

When there is no outcome favorable, i.e., the number of outcomes is zero.

The probability of an event that is certain to happen is 1. Such an event is called a sure/certain event

When the number of favorable outcomes is the same as the number of all possible outcomes it is a sure event.

The sum of the probabilities of all the elementary events of an experiment is 1 .

The probability of an event is greater than or equal to 0 and less than or equal to 1 .

It is not an equally likely event since it relies on various factors that are not alike for both the outcomes.

A player attempts to shoot a basketball. She/he shoots or misses the shot.

It is not an equally likely event, because it depends on the ability and amount of practice of the player. If he is a professional player, he will more likely have a successful shot. Whereas an amateur player will more likely miss the shot.

It is an equally likely event. The only options are true or false and only one of them is correct.

A baby is born. It is a boy or a girl.

It is an equally likely event. The only possibilities of gender are boy and girl. Hence if not boy then girl and vice versa.

The tossing of the coin is considered to be a fair way of deciding because the only possible outcomes are head and tails. Hence they are equally likely events.

(A) 2/3

(B) –1.5

(C) 15%

(D) 0.7

We know, probability of an event is either greater than or equal to 0 and always less than or equal to 1. Hence the probability of an event can never be negative.

Therefore, (B) $-1.5$ cannot be the probability of an event.

Also, (A) : $\dpi{100} \frac{2}{3} = 0.67$

(C): $\dpi{100} 15\% = \frac{15}{100} = 0.15$

(D): 0.7

Hence (A), (C), (D) all lie between 0 and 1.

Given, $P(E) = 0.05$

We know,

$P(not\ E) = 1 - P(E)$

$\therefore P(not\ E) = 1 - 0.05 = 0.95$

Hence, the probability of 'not E' is 0.95

According to the question, the bag contains only lemon-flavored candies. It does not contain any orange flavor candy. Hence, every time only lemon flavor candy will come out. Therefore, $P(an\ orange\ flavoured\ candy) = 0$ i.e. event of taking out an orange-flavored candy is an impossible event.

According to the question, the bag contains only lemon-flavored candies. So the event that Malini takes out a lemon-flavored candy is a sure event. Therefore, $P(a\ lemon\ flavoured\ candy) = 1$

Given,

Probability of two students not having the same birthday $P (\overline E) = 0.992$

$\therefore$ Probability of two students having the same birthday = $P (E) = 1 - P (\overline E)$

$= 1 - 0.992 = 0.008$

Hence, the probability that the 2 students have the same birthday is 0.008

Total number of balls in the bag = 8
No. of red balls = 3
No. of black balls = 5

(i) Let E be the event of getting a red ball

n(E) = No. of red balls = 3
n(S) = No. of total balls = 8

$\therefore$ Probability of the ball drawn to be red =
$P = \frac{n(E)}{n(S)}$
$= \frac{3}{8}$

Total number of balls in the bag = 8
No. of red balls = 3
No. of black balls = 5

(ii) We know,
$P(not\ E) = P (\overline E) = 1 - P (E)$
where $\dpi{100} E\ and\ \overline E$ are complementary events.

$\therefore$ Probability of not getting the red ball
$\dpi{100} = 1 - Probability\ of\ getting\ a\ red\ ball$
$\dpi{100} \\ = 1 - \frac{3}{8} \\ = \frac{5}{8}$

Given,

Total number of balls in the bag = 5 + 8 + 4 = 17

(i) Let R be the event that the ball taken out is red

The number of possible outcomes = 17

The number of outcomes favorable to the event R = 5

$\therefore$ $P(R) = \frac{favourable\ outcomes}{total\ outcomes}=\frac{5}{17}$

Given,

Total number of balls in the bag = 5 + 8 + 4 = 17

(ii) Let W be the event that the ball taken out is white

The number of possible outcomes = 17

The number of outcomes favorable to the event W = 8

$\therefore$ $P(W) = \frac{favourable\ outcomes}{total\ outcomes}=\frac{8}{17}$

Given,

Total number of balls in the bag = 5 + 8 + 4 = 17

(iii) Let G be the event that the ball taken out is green

The number of possible outcomes = 17

The number of outcomes favorable to the event G = 4

$\therefore$ $P(G) = \frac{favourable\ outcomes}{total\ outcomes}=\frac{4}{17}$

$\\ \therefore P(not\ G) =P(\overline G) =1-P(G) \\ = 1-\frac{4}{17} \\ = \frac{13}{17}$

The required probability of not getting a green ball is $\frac{13}{17}$

Total number of coins in the piggy bank = 100+50+20+10 = 180

Let E be the event of getting a 50p coin.

Number of possible outcomes = 180

Number of outcomes favorable to event E = 100

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{100}{180}$

$= \frac{5}{9}$

Therefore, the probability of getting a 50p coin is $\dpi{100} \frac{5}{9}$

Total number of coins in the piggy bank = 100+50+20+10 = 180

Let F be the event of getting an Rs. 5 coin.

Number of possible outcomes = 180

Number of outcomes favorable to event E = 10

$\therefore P(F) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{10}{180}$

$= \frac{1}{18}$

$\therefore P(not\ getting\ a\ Rs.\ 5\ coin )=P(\overline F)$

$= 1 - P(F) = 1 - \frac{1}{18} = \frac{17}{18}$

Therefore, the probability of not getting an Rs. 5 coin is $\dpi{100} \frac{17}{18}$

Total number of fishes in the tank = 5 (male) + 8 (female) = 13

Let E be the event that the fish taken out is a male fish.

Number of possible outcomes = 13

Number of outcomes favorable to E = 5

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{5}{13}$

Therefore, the probability that the fish are taken out is a male fish is $\dpi{100} \frac{5}{13}$

Total posible outcomes = {1,2,3,4,5,6,7,8}

Number of possible outcomes = 8

Let E be the event of getting 8.

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{1}{8}$

Therefore, the probability that it will point at 8 is $\dpi{100} \frac{1}{8}$

Total posible outcomes = {1,2,3,4,5,6,7,8}

Number of possible outcomes = 8

Let E be the event of pointing at an odd number.

Total number of odd numbers = n({1,3,5,7}) = 4

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{4}{8}$

$= \frac{1}{2}$
Therefore, the probability of getting an odd number is $\dpi{100} \frac{1}{2}$

Total posible outcomes = {1,2,3,4,5,6,7,8}

Number of possible outcomes = 8

Let E be the event of pointing at number greater than 2

Number of favouable outcomes= n({3,4,5,6,7}) = 5

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{5}{8}$
Therefore, the probability of pointing at a number greater than 2 is $\dpi{100} \frac{5}{8}$

Total possible outcomes = {1,2,3,4,5,6,7,8}

Number of possible outcomes = 8

Let E be the event of pointing at a number less than 9

Since all the numbers on the wheel are less than 9, this is the sure event.

Number of favorable outcomes = 8

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{8}{8} = 1$
Therefore, the probability of pointing at a number less than 9 is $\dpi{100} 1$
.

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Number of possible outcomes once = 6

(i) Let E be the event of getting a prime number.

Prime numbers on the die are = {2,3,5}

Number of favorable outcomes = 3

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{3}{6}$

$= \frac{1}{2}$

Therefore, the probability of getting a prime number is $\dpi{80} \dpi{100} \frac{1}{2}$

Possible outcomes when a die is thrown once = {1,2,3,4,5,6}

Number of possible outcomes = 6

(ii) Let F be the event of getting a number lying between 2 and 6

Numbers lying between 2 and 6 on the die are = {3,4,5}

Number of favorable outcomes = 3

$\therefore P(F) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{3}{6}$

$= \frac{1}{2}$

Therefore, the probability of getting a number lying between 2 and 6 is $\dpi{100} \frac{1}{2}$

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Number of possible outcomes once = 6

(iii) Let O be the event of getting an odd number.

Odd numbers on the die are = {1,3,5}

Number of favorable outcomes = 3

$\therefore P(O) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{3}{6}$

$= \frac{1}{2}$

Therefore, the probability of getting an odd number is $\dpi{100} \frac{1}{2}$ .

Total number of cards in a well-shuffled deck = 52

Hence, total possible outcomes = 52

(1) Let E be the event of getting a king of red color.

There are only red color kings: Hearts and diamonds

Hence, number of favorable outcomes = 2

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{2}{52}$

$= \frac{1}{26}$

Therefore, the probability of getting a king of red color is $\dpi{80} \frac{1}{26}$

Total number of cards in a well-shuffled deck = 52

Hence, total possible outcomes = 52

(2) Let E be the event of getting a face card.

Face cards: (J, Q, K) of each four suits

Hence, number of favorable outcomes = 12

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{12}{52}$

$= \frac{3}{13}$

Therefore, the probability of getting a face card is $\dpi{80} \frac{3}{13}$

Total number of cards in a well-shuffled deck = 52

Hence, total possible outcomes = 52

(3) Let E be the event of getting a red face card.

Face cards: (J, Q, K) of hearts and diamonds

Hence, number of favourable outcomes = 3x2 = 6

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{6}{52}$

$= \frac{3}{26}$

Therefore, the probability of getting a red face card is $\dpi{80} \frac{3}{26}$

Total number of cards in a well-shuffled deck = 52

Hence, total possible outcomes = 52

(4) Let E be the event of getting the jack of hearts

Hence, the number of favourable outcomes = 1

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{1}{52}$

Therefore, the probability of getting the jack of hearts is $\dpi{80} \frac{1}{52}$

Total number of cards in a well-shuffled deck = 52

Hence, total possible outcomes = 52

(5) Let E be the event of getting a spade.

There are 13 cards in each suit. {2,3,4,5,6,7,8,9,10,J,Q,K,A}

Hence, number of favourable outcomes = 13

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{13}{52}$

$= \frac{1}{4}$

Therefore, the probability of getting a spade is $\dpi{80} \frac{1}{4}$

Total number of cards in a well-shuffled deck = 52

Hence, total possible outcomes = 52

(6) Let E be the event of getting the queen of diamonds

Hence, the number of favorable outcomes = 1

$\therefore P(E) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{1}{52}$

Therefore, the probability of getting the queen of diamonds is $\dpi{80} \frac{1}{52}$

Total number of cards = 5

Hence, the total possible outcomes = 5

(1) There is only one queen.

Hence, favorable outcome = 1

$\therefore P(getting\ a\ queen) = \frac{1}{5}$

When the queen is kept aside, there are only 4 cards left

Hence, the total possible outcomes = 4

(2a) There is only one ace.

Hence, favorable outcome = 1

$\therefore P(getting\ an\ ace)= \frac{1}{4}$

Therefore, the probability of getting an ace is 0.25

When the queen is kept aside, there are only 4 cards left

Hence, the total possible outcomes = 4

(2b) Since there is no queen left.

Hence, favorable outcome = 0

$\therefore P(getting\ a\ queen)= \frac{0}{4} = 0$

Therefore, the probability of getting a queen is 0. Thus, it is an impossible event.

Total number of pens = 132(good) + 12(defective)

Hence, the total possible outcomes = 144

Number of good pens = number of favorable outcomes = 132

$\therefore P(getting\ a\ good\ pen) = \frac{favourable\ outcome}{total\ outcome} = \frac{132}{144} = \frac{11}{12}$

Total number of bulbs = 20

Hence, total possible outcomes = 20

Number of defective bulbs = 4

Hence, the number of favorable outcomes = 4

$\therefore P(getting\ a\ defective\ bulb)= \frac{favourable\ outcomes}{total\ outcomes} = \frac{4}{20}$

$= \frac{1}{5}$

Total number of bulbs = 20

Hence, total possible outcomes = 20

Number of defective bulbs = 4

Hence, the number of favorable outcomes = 4

$\therefore P(getting\ a\ defective\ bulb)= \frac{favourable\ outcomes}{total\ outcomes} = \frac{4}{20}$

$= \frac{1}{5}$

$\therefore P(getting\ a\ non\ defective\ bulb)= 1 - \frac{1}{5} = \frac{4}{5}$

Total number of discs = 90

Number of discs having a two-digit number between 1 and 90 = 81

$\therefore P(getting\ a\ two-digit\ number) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{81}{90}$

$= \frac{9}{10}$

Total number of discs = 90

Perfect square numbers between 1 and 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81}

Therefore, the total number of discs having perfect squares = 9.

$\therefore P(getting\ a\ perfect\ square) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{9}{90}$

$= \frac{1}{10}$

Total number of discs = 90

Numbers between 1 and 90 that are divisible by 5 are {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90}

Therefore, total number of discs having numbers that are divisible by 5 = 18.

$\therefore P(getting\ a\ number\ divisible\ by\ 5) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{18}{90}$

$= \frac{1}{5}$

The die is thrown once. What is the probability of getting (i) A?

The six faces of the die contains : {A,B,C,D,E,A}

Total number of letters = 6

(i) Since there are two A's,

number of favorable outcomes = 2

$\therefore P(getting\ A) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{2}{6}$

$= \frac{1}{3}$

Therefore, the probability of getting A is $\dpi{100} \frac{1}{3}$

The die is thrown once. What is the probability of getting (ii) D?

The six faces of the die contains : {A,B,C,D,E,A}

Total number of letters = 6

(i) Since there is only one D,

number of favorable outcomes = 1

$\therefore P(getting\ A) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{1}{6}$

Therefore, the probability of getting D is $\dpi{100} \frac{1}{6}$

Here, the Total outcome in the area of the rectangle and favorable outcome is the area of the circle.

Area of the rectangle = $l\times b = 3\times2 = 6\ m^2$

Area of the circle = $\pi r^2 = \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{4}\ m^2$

$\therefore P(die\ will\ land\ inside\ the\ hole) = \frac{Area\ of\ circle}{Total\ area}$

$= \frac{\frac{\pi}{4}}{6} = \frac{\pi}{24}$

Total number of pens = 144

Total number of defective pens = 20

$\therefore$ Number of good pens = 144-20 = 124

She will buy it if the pen is good.

Therefore, the probability that she buys = probability that the pen is good =

$P(getting\ a\ good\ pen) = \frac{number\ of\ good\ pens}{total\ pens}= \frac{124}{144}$

$= \frac{31}{36}$

Total number of pens = 144

Total number of defective pens = 20

She will buy it if the pen is good.

Therefore, the probability that she will not buy = probability that the pen is defective =

$P(getting\ a\ defective\ pen) = \frac{no.\ of\ defective\ pens}{total\ pens}= \frac{20}{144}$

$= \frac{5}{36}$

 Event: 'sum on 2 dice' 2 3 4 5 6 7 8 9 10 11 12 1/36 5/36 1/36

The table becomes:

 The sum of two dice 2 3 4 5 6 7 8 9 10 11 12 Probability 1/36 1/18 1/12 1/9 5/36 1/6 5/36 1/9 1/12 1/18 1/36

A student argues that "there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability of 1/11. We do not agree with this argument because there are a different number of possible outcomes for each sum. we can see that each sum has a different probability.

The possible outcomes when a coin is tossed 3 times: (Same as 3 coins tossed at once!)

{HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Number of total possible outcomes = 8

For Hanif to win, there are only two favorable outcomes: {HHH, TTT}

Number of favorable outcomes = 2

$\therefore P(Hanif\ will\ win) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{2}{8}$

$= \frac{1}{4}$

$\therefore P(Hanif\ will\ lose) = 1-\frac{1}{4}$

$=\frac{3}{4}$

Therefore, the probability that Hanif will lose is $\frac{3}{4}$

When a die is thrown twice, the possible outcomes =

$\{(x,y): x,y\in\{1,2,3,4,5,6\}\}$

Total number of possible outcomes = $6\times6 = 36$

The outcomes when 5 comes up either on them =

{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}

Number of such favorable outcomes = 11

$\therefore P(5\ will\ come\ up\ either\ time) = \frac{11}{36}$

$\therefore P(5\ will\ not\ come\ up\ either\ time) = 1-\frac{11}{36} =\frac{25}{36}$

Therefore, the probability that 5 will not come either time is $\frac{25}{36}$

When a die is thrown twice, the possible outcomes =

$\{(x,y): x,y\in\{1,2,3,4,5,6\}\}$

Total number of possible outcomes = $6\times6 = 36$

The outcomes when 5 comes up at least once =

{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}

Number of such favorable outcomes = 11

$\therefore P(5\ comes\ up\ at\ least\ once)= \frac{11}{36}$

Therefore, the probability that 5 comes at least once is $\frac{11}{36}$

The possible outcomes when two coins are tossed = {HH, HT, TH, TT}

Total number of possible outcomes = 4

$\therefore P(getting\ two\ heads) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{1}{4}$

Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence the mentioned events are not equally likely.

The possible outcomes when a die is thrown= {1,2,3,4,5,6}

Total number of possible outcomes = 6

Number of odd number, {1,3,5} = 3

And, number of even numbers {2,4,6} = 3

Hence, both these events are equally likely

$\therefore P(getting\ an\ odd) = \frac{favourable\ outcomes}{total\ outcomes} = \frac{3}{6} = \frac{1}{2}$

More About NCERT Solutions for Class 10 Maths Exercise 15.1 –

NCERT solutions Class 10 Maths exercise 15.1 is based on the likelihood of something or some particular event to occur. The basis of experimental probability is directly based on the observations of any experiment conducted in random. It can be calculated by dividing the total number of trials by the number of possible outcomes. For example, if a coin is tossed eight times and heads are recorded six times, the experimental probability for heads is.

Also Read| Probability Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 15.1

• Exercise 15.1 Class 10 Maths, is based on probability and the implication of probability.

• NCERT syllabus Class 10 Maths chapter 15 exercise 15.1 helps us to grasp the basic concept of probability by solving some of the basic questions related to it.

• Class 10 Maths chapter 15 exercise 15.1 prepares us for the new types of problem which are to come in the next exercise

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JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

NCERT Solutions for Class 10 Subject Wise

1. What is the core concept behind NCERT solutions for Class 10 Maths exercise 15.1?

The core concept of this exercise deals with probability. Probability of occurrence of an event, formula for its calculation, basics terminology and its conditions.

2. Define probability in simple terms?

In general terms probability is the ratio of total number of favourable outcomes for an event to the total number of outcomes present in the sample space of the experiment.

3. What is meant by sample space?

The collection of all possible outcomes for an event is called sample space for an event.

4. Can probability of an event be negative?

No, probability is only for true events and it cannot be negative.

5. A coin is tossed randomly then the Probability of head to occur is?

No, probability is only for true events and it cannot be negative.

6. A bag contains 10 balls of which 3 are red and 7 are black. We pick a ball from the bag, the Probability of red to occur is?

No, probability is only for true events and it cannot be negative.

7. Number of possible outcomes If we Roll a dice is?

So the total outcomes = {1, 2, 3, 4, 5, 6}

Total no of outcomes = 6

8. Probability of an event to occur is 0.46 then the Probability that the event doesn’t occur is?

Probability of the event = 0.46

Probability that the event doesn’t occur = 1 – 0.46 = 0.54

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

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Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9