Finding The C.P. When S.P. and Profit Or Loss Percent Are Given

Formulas

Profit % = (\(\frac{ Profit}{ C.P.}\) ) x 100

Loss % = (\(\frac{ Loss}{ C.P.}\)) x 100

C.P. = \(\frac{ 100 X S.P.}{ (100 – Loss \%)}\)

C.P. = \(\frac{ 100 X S.P.}{ (100 + Profit \%)}\)

Illustrative Examples

Example 1: By selling a fan for Rs 649, Anil earns a profit of 18%. Find its cost price.

Solution. S.P. of the fan = Rs 649, profit = 18%

Therefore, Rs 649 = ( 1 + \(\frac{18}{100}\)) of C.P.

=> Rs 649 = \(\frac{118}{100}\) of C.P.

=>  C.P. = Rs (649 x \(\frac{100}{118}\)) = Rs 550
Hence the cost price of the fan = Rs 550.

Example 2: By selling a chair for Rs 391, Ali suffers a loss of 15%. Find its cost price.

Solution. S.P. of the chair = Rs 391, Loss = 15%

Therefore,    Rs 391 = ( 1- \(\frac{15}{100}\)) of C.P.

Rs 391 = \(\frac{85}{100}\) of C.P.

=>  C.P. = (Rs 391 x \(\frac{100}{85}\) )= Rs (23 x 20) = Rs 460

Hence the cost price of the chair = Rs 460.

Example 3: A man sells his scooter for Rs 18000 making a profit of 20%. How much did the scooter cost him?

Solution. Let the cost price of the scooter be Rs 100. Then, Profit = Rs 20

S.P. = C.P. + Profit = Rs 100 + Rs 20= Rs 120
Thus, if the S.P. is Rs 120, then C.P. = Rs 120 – 20 = Rs 100

If the S.P. is Rs 18000, then C.P. = Rs (\(\frac{100}{120}\) x 18000) = Rs 15000

Hence, the cost of the scooter = Rs 15000

Decimal Number System

Introduction

This system is based upon the place value and face value of a digit in a number. We have learnt that a natural number can be written as the sum of the place values of all digits of the numbers. For example

3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1
Such a form of a natural number is known as its expanded form.

The expanded form of a number can also be expressed in terms of powers of 10 by using

\( {10}^{0} \) = 1, \( {10}^{1} \) = 10, \( {10}^{2} \) = 100, \( {10}^{3} \) = 1000 etc.
For example,

3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1
=>    3256 = 3 x \( {10}^{3} \) + 2 x \( {10}^{2} \) + 5 x \( {10}^{1} \) + 6 x \( {10}^{0} \)

Clearly, each digit of the natural number is multiplied by \( {10}^{n} \), where n is the number of digits to its right and then they are added.

Illustrative Examples

Example 1:    Write the following numbers in the expanded exponential forms:

(i) 32005    (ii) 56719    (iii) 8605192    (iv) 2500132
Solution.

(i) 32005 = 3 x \( {10}^{4} \) + 2 x \( {10}^{3} \) + 0 x \( {10}^{2} \) + 0 x \( {10}^{1} \) + 5 x \( {10}^{0} \)

(ii) 560719 = 5 x \( {10}^{5} \) + 6 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 7 x \( {10}^{2} \) + 1 x \( {10}^{1} \) + 9 x \( {10}^{0} \)

(iii) 8605192 = 8 x \( {10}^{6} \) + 6 x \( {10}^{5} \) + 0 x \( {10}^{4} \) + 5 x \( {10}^{3} \) + 1 x \( {10}^{2} \) + 9 x \( {10}^{1} \) + 2 x \( {10}^{0} \)

(iv) 2500132 = 2 x \( {10}^{6} \) + 5 x \( {10}^{5} \) + 0 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 1 x \( {10}^{2} \) + 3 x \( {10}^{1} \) + 2 x \( {10}^{0} \)

Example 2:    Find the number from each of the following expanded forms:

(i) 5 x \( {10}^{4} \) + 4 x \( {10}^{3} \) + 2 x \( {10}^{2} \) + 3 x \( {10}^{1} \) + 5 x \( {10}^{0} \)

(ii) 7 x \( {10}^{5} \) + 6 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 9 x \( {10}^{0} \)

(iii) 9 x \( {10}^{5} \) + 4 x \( {10}^{2} \) + 1 x \( {10}^{1} \)

Solution.

(i) 5 x \( {10}^{4} \) + 4 x \( {10}^{3} \) + 2 x \( {10}^{2} \) + 3 x \( {10}^{1} \) + 5 x \( {10}^{0} \)

= 5 x 10000 + 4 x 1000 + 2 x 100 + 3 x 10 + 5 x 1

= 50000 + 4000 + 200 + 30 + 5

= 54235

(ii) 7 x \( {10}^{5} \) + 6 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 9 x \( {10}^{0} \)

=  7 x 100000 + 6 x 10000 + 0 + 9 x 1

= 700000 + 60000 + 9

= 760009

(iii) 9 x \( {10}^{5} \) + 4 x \( {10}^{2} \) + 1 x \( {10}^{1} \)

= 9 x 100000 + 4 x 100 + 1 x 10

= 900000 + 400 + 10

= 900410

Posted in Maths  • 

Compound Interest

Definitions

Compound Interest: If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half year or a quarter of a year etc) so that the amount ( = Principal + Interest) at the end of an interval becomes the principal for the next interval, then the total interest over all the intervals, calculated in this way is called the compound interest and is abbreviated as C.I.

Clearly, compound interest at the end of certain specified period is equal to the difference between the amount at the end of the period and the original principal i.e.
C.I. = Amount — Principal

Conversion Period: The fixed interval of time at the end of which the interest is calculated and added to the principal at the beginning of the interval is called the conversion period.

In other words, the period at the end of which the interest is compounded is called the conversion period.

When the interest is calculated and added to the principal every six months, the conversion period is six months. Similarly, the conversion period is 3 months when the interest is calculated and added quarterly.

Finding CI When Interest Is Compounded Annually

when Interest is compounded yearly, the interest accrued during the first year is added to the principal and the amount so obtained becomes the principal for the second year. The amount at the end of the second year becomes the principal for the third year, and so on.

Example 1: Maria invests Rs 93750 at 9.6% per annum for 3 years and the interest is compounded annually. Calculate:
(i) The amount standing to her credit at the end of second year.
(ii) The interest for the third year.
Solution.    (i) We have,
Principal for the first year Rs 93750

Rate of interest = 9.6% per annum.

Therefore,          Interest for the first year = Rs (\(\frac{93750 \times 9.6 \times 1}{100}\)) = Rs 9000

Amount at the end of the first year = Rs 93750 +Rs 9000

= Rs 102750

Principal for the second year = Rs 102750

Interest for the second year = Rs (\(\frac{102750 \times 9.6 \times 1}{100}\)) = Rs 9864

Amount at the end of second year = Rs 102750 + Rs 9864

= Rs 112614

(ii) Principal for the third year = Rs 112614

Interest for the third year =Rs (\(\frac{112614 \times 9.6 \times 1}{100}\) ) = Rs 10810.94

Example 2: Find the compound interest on Rs 25000 for 3 years at 10% per annum, compounded annually.

Solution.    Principal for the first year = Rs 25000

Interest for the first year  = (\(\frac{25000 \times 10 \times 1}{100}\)) = Rs 2500

Amount at the end of the first year = (25000 + 2500) = Rs 27500

Principal for the second year = Rs 27500

Interest for the second year = (\(\frac{27500 \times 10 \times 1}{100}\)) = Rs 2750

Amount at the end of the second year = (27500 + 2750) = Rs 30250.

Principal for the third year = Rs 30250.

Interest for the third year = (\(\frac{30250 \times 10 \times 1}{100}\)) = Rs 3025

Amount at the end of the third year = (30250 + 3025) = Rs 33275.

Therefore, compound interest = (33275 — 25000) = Rs 8275

Finding CI When Interest Is Compounded Half-Yearly

If the rate of Interest is R% per annum then it is clearly (\(\frac{R}{2}\))% per half-year.
The amount after the first half-year becomes the principal for the next half-year, and so on.

Example 3: Find the compound interest on Rs 5000 for 1 year at 8% per annum, compound half-yearly.

Solution.    Rate of interest             = 8% per annum = 4% per half-year.

Time                                          = 1 year = 2 half-years.

Original principal                    = Rs 5000.

Interest for the first half-year = (\(\frac{5000 \times 4 \times 1}{100}\)) = Rs 200.

Amount at the end of the first half-year = (5000 + 200) = 5200.
Principal for the second half-year = Rs 5200.

Interest for the second half-year = ((\(\frac{5200 \times 4 \times 1}{100}\))) = Rs 208.

Amount at the end of the second half-year = Rs (5200 + 208) = Rs 5408.
Therefore,        compound interest = Rs (5408 — 5000) = Rs 408.

Example 4: Find the compound interest on Rs 8000 for \(1\frac{1}{2}\) years at 10% per annum, interest being payable half-yearly.

Solution.    We have,

   Rate of interest                = 10% per annum = 5% per half-year.

   Time                                  = \(1\frac{1}{2}\) years = \(\frac{3}{2}\) x 2 = 3 half- years

Original principal             = Rs 8000

Interest for the first half-year = Rs ( \(\frac{8000 \times 5 \times 1}{100}\)) = Rs 400

Amount at the end of the first half-year = Rs 8000 + R 400 = Rs 8400
Principal for the second half-year = Rs 8400

Interest for the second half-year = Rs (\(\frac{8400 \times 5 \times 1}{100}\)) = Rs 420

Amount at the end of the second half-year = Rs 8400 + Rs 420 = Rs 8820
Principal for the third half-year = Rs 8820

Interest for the third half-year = Rs (\(\frac{8820 \times 5 \times 1}{100}\)) = Rs 441

Amount at the end of third half-year = Rs 8820 + Rs 441 = Rs 9261
Therefore,        Compound interest = Rs 9261 — Rs 8000 = Rs 1261

Finding CI When Interest Is Compounded Quarterly

If the rate of interest is R % per annum and the interest is compounded quarterly, then it is \(\frac{R}{4}\) %  per quarter.

Example 5: Find the compound interest on Rs 10000 for 1 year at 20% per annum compounded quarterly.

Solution.    We have,

Rate of interest             = 20% per annum = \(\frac{1}{5}\)% = 5% per quarter

Time                         = 1 year = 4 quarters.

Principal for the first quarter = Rs 10000

Interest for the first quarter = Rs (\(\frac{10000 \times 5 \times 1}{100}\)) = Rs 500

Amount at the end of first quarter = Rs 10000 + Rs 500 = Rs 10500

Principal for the second quarter = Rs 10500

Interest for the second quarter = Rs (\(\frac{10500 \times 5 \times 1}{100}\)) = Rs 525

Amount at the end of second quarter = Rs 10500 + Rs 525 = Rs 11025
Principal for the third quarter = Rs 11025

Interest for the third quarter = Rs (\(\frac{11025 \times 5 \times 1}{100}\)) = Rs 551.25

Amount at the end of the third quarter = Rs 11025 + Rs 551.25

= Rs 11576.25

Principal for the fourth quarter = Rs 11576.25

Interest for the fourth quarter = Rs (\(\frac{11576.25 \times 5 \times 1}{100}\)) = Rs 578.8125

Amount at the end of fourth quarter = Rs 11576.25 + Rs 578.8125

= Rs 12155.0625

Therefore,        Compound interest = Rs 12155.0625 — Rs 10000

= Rs 2155.0625

= Rs 2155.06

Posted in Maths  • 

Floatation- Textbook Exercises

Q1. In what direction does the buoyant force on an object immersed in a liquid act?

Ans. The buoyant force on an object immersed in a liquid acts in upward direction (i.e., opposite to weight of the object).

Q2. Why does a block of plastic released under water come up to the surface of water?

Ans. The density of plastic is less than that of water, so the force of buoyancy on plastic block(V\(\rho\)plastic g) will be greater than the weight of plastic block displaced, i.e., (V \(\rho\) water  g). Hence, the acceleration of plastic block will be in upward direction, and comes up to the surface of water.
Mathematically,
acceleration, a = B — W

= \(\frac { V\quad { \rho  }_{ water }\quad g\quad -\quad V\quad { \rho  }_{ plastic }\quad g }{ m } \)

= \(\frac { V\quad g\quad ({ \rho  }_{ water }\quad \quad -\quad \quad { \rho  }_{ plastic })\quad  }{ m } \)

is positive in upward direction.

Q3. The volume of 50 g of a substance is 20cm . If the density of water is 1 g cm-3 , will the substance float or sink?

Ans. Density of substance,  \(\rho \) =\(\frac{mass}{volume}\) = \(\frac { 50\quad g }{ 20\quad { cm }^{ 3 } } \) = 2.5 g cm-3
As density of substance is greater than the density of water, therefore, the substance will sink in water. Alternatively, the buoyant force exerted by water (\((V\quad { \rho  }_{ water }\quad g)\)) on substance, when fully immersed, is less than the weight of substance, i.e., B< W; hence, the substance will sink in water.

Q4. The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm-3 ? What will be the mass of the water displaced by this packet?

Ans. Density of packet, \(\rho =\frac { mass }{ volume } \)

=\(\frac { 500\quad g }{ 350\quad { cm }^{ 3 } } \)

=\(\frac { 10\quad  }{ 7 } g\quad { cm }^{ -3 }\)

= l.43 g cm-3
As density of packet is more than the density of water, so the packet will sink in water.
Weight of water displaced by the packet
= volume of packet x density of water
= (350 cm3) x (l g cm-3) = 350 g
Obviously, the force of buoyancy, B = 350 g wt and weight of packet, W = 500 g wt i.e., B < W; so the packet sinks in water.

Floatation–Textbook Questions

Q1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

If a school bag has a strap made of thin and strong string, the whole weight of school bag will fall of over a very small area (equal to that string) on the shoulder. So pressure on shoulder P =\(\frac{Weight of the bag}{Area of string}\) of child will be very large and hence painful for the child. Hence, it is difficult to hold a school bag having a strap made of thin string.

Q2. What do you mean by buoyancy?

Ans. When an object is placed in a liquid, the liquid exerts an upward force on it. The tendency of liquid to exert such an upward force on the object is called the buoyancy and the upward force exerted by liquid on the object is called the buoyant force (or the force of buoyancy).

Q3. Why does an object float or sink when placed over the surface of water?

Ans. When an object is immersed in water, the water exerts an upward force on the object. This upward force equal to the weight of water displaced by object is called the buoyant Force.
If on completely immersing the object, the buoyant force is more than the weight of object will float on water. In other words, if the density of object is less than that of water, then the object will float on water.
On the other hand if buoyant force, on completely immersing the object in water, is less than the weight of the object, the object will sink in water. In other words, if the density of object is greater than water, then the object will sink in water.
In brief, if  p object < p water   , the object will float and if p object > p water, the object will sink.

Q4. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

Ans. When we stand on a weighing machine, we replace air equal to volume of our body. So measured weight, W = actual weight — upthrust
= mg — V p air g
Therefore, Measured mass, m’ = \(\frac { W }{ g }\) = (m-V air )
That is the measured mass will be less than the actual mass. Therefore, our mass is more than 42 kg.

Q5. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

Ans. Actually mass of cotton is more than the mass of iron bar. This is because
Weight measured = Actual weight — buoyant force
Therefore,  Actual weight =  weight measured + buoyant force
The volume of cotton is much more than the volume of iron bar, so buoyant force (B = volume of object x density of air x g) of air on cotton is much more than that of air on iron bar, so mass of cotton will be more than the mass of iron bar.

Exponents

Definition

If a is any real number and n is a natural number, then \( { a }^{ n } \) = a x a x a…. n times

where a is called the base, n is called the exponent or index and  \( { a }^{ n } \) is the exponential expression. \( { a }^{ n } \) is read as ‘a raised to the power n’ or ‘a to the power n’ or simply ‘a power n’.

For zero power, we have :

\( { a }^{ 0 } \) = 1 (where a \( \neq \) 0)

For example :

(i) \( { 7 }^{ 0 } \) =  1    (ii) \( { (-\frac { 2 }{ 3 } ) }^{ 0 } \) = 1    (iii) \(( { \sqrt { 7 }  })^{ 0 } \) = 1

For negative powers, we have :

\( \sqrt [ n ]{ { a }} \) = \( \frac { 1 }{ { a }^{ n } } \) and \( \frac { 1 }{ { a }^{ -n } } \) = \( { a }^{ n } \)

For example:
(i) \( { 5 }^{ -2 } \) = \( \frac { 1 }{ { 5 }^{ 2 } } \)
(ii) \( { -2 }^{ -3 } \) = \( \frac { 1 }{ { -2 }^{ 3 } } \)
(iii) \( \frac { 1 }{ { 2 }^{ -5 } } \) = \( { 2 }^{ 5 } \)

For fractional indices,  we have :

\( { \sqrt { a}  }^{ n } \) = \( { a }^{ \frac { 1 }{ n }  } \) and \( \sqrt [ n ]{ { a }^{ m } } \) = \( { a }^{ \frac { m }{ n }  } \)

For example:

(i) \( { \sqrt { 3 }  }\) = \( { 3 }^{ \frac { 1 }{ 2 }  } \)
(ii) \( { \sqrt { 8 }  }^{ 3 } \) = \( { 8 }^{ \frac { 1 }{ 3}  } \)
(iii) \( \sqrt [ 4 ]{ { 5 }^{ 3 } } \) = \( { 5 }^{ \frac { 3 }{ 4 }  } \)

Finding the value of the Number given in the Exponential Form

Example 1: Find the value of each of the following:

(i) \( { 12 }^{ 2 } \)    (ii) \( { 8 }^{ 3 } \)    (iii) \( { 4 }^{ 4 } \)

Solution.

(i) We have,

\( { 12 }^{ 2 } \) = 12 x 12 = 144
(ii) We have,

\( { 8 }^{ 3 } \) = 8 x 8 x 8

= (8 x 8) x 8
= 64 x 8
= 512
(iii) We Have,

\( { 4 }^{ 4 } \)= 4 x 4 x 4 x 4

= (4 x 4 ) x 4 x 4
= (16 x 4) x 4
= 64 x 4
= 256

Example 2: Simplify:

(i) 2 x \( { 10 }^{ 3 } \)    (ii) \( { 5 }^{ 2 } \) x \( { 4 }^{ 2 } \)    (iii) \( { 3 }^{ 3 } \) x 4

Solution.

(i) We have,

2 x \( { 10 }^{ 3 } \) = 2 x 1000 = 2000     [since \( { 10 }^{ 3 } \)=10 x10 x 10 = 1000]

(ii) We have,

\( { 5 }^{ 2 } \) x \( { 4 }^{ 2 } \)

= 25 x 16 = 400

(iii) We have,

\( { 3 }^{ 3 } \) x 4  = 27 x 4 = 108

Expressing Numbers in Exponential Form

Example 1: Express each of the following in exponential form:

(i) (-4) x (-4) x (-4) x (-4) x (-4)    (ii) \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\)
Solution. We have,

(i)  (-4) x (-4) x (-4) x (-4) x (-4) = \( { -4}^{ 5 } \)

(ii) \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\) = \( ({ \frac { 2 }{ 5 } ) }^{ 4 } \)

Example 2: Express each of the following in exponential form:

(i) 3 x 3 x 3 x a x a    (ii) a x a x a x a x a x a x b x b x b c x c x c x c
(iii) b x b x b x \(\frac{2}{5}\) x \(\frac{2}{5}\)
Solution. We have,

(i) 3 x 3 x 3 x a x a = \( { 3 }^{ 3 } \) x \( { a }^{ 2 } \)

(ii) a x a x a x a x a x a x b x b x b x c x c x c x c = \( { a }^{ 6 } \) x \( { b }^{ 3 } \) x \( { c }^{ 4 } \)

(iii) b x b x b x \(\frac{2}{5}\) x \(\frac{2}{5}\) = \( { a }^{ 3 } \) x \( ({ \frac { 2 }{ 5 } ) }^{ 2 } \)

Example 3: Express each of the following numbers in exponential form:

(i) 128    (ii) 243    (iii) 3125
Solution.

(i) We have,

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
128 = \( { 2 }^{ 7 } \)

Express-In-Exponential-Form-example3(i)

(ii) We have,

243 = 3 x 3 x 3 x 3 x 3
243 = \( { 3 }^{ 5 } \)

Express-In-Exponential-Form-example3(ii)
(iii) We have,

625 = 5 x 5 x 5 x 5

625 = \( { 5 }^{ 4 } \)

Express-In-Exponential-Form-example3(iii)

Positive Integral Exponent of a Rational Number

Let \(\frac{a}{b}\) be any rational number and n be a positive integer. Then,

\( {(\frac { a } { b })^{ n } } \) = \(\frac{a}{b}\) x \(\frac{a}{b}\) x \(\frac{a}{b}\)…n times
= \( \frac { a\quad \times \quad a\quad \times \quad a….n\quad times }{ b\quad \times \quad b\quad \times \quad b….n\quad times } \)
= \( \frac { { a }^{ n } }{ { b }^{ n } } \)
Thus, \( {(\frac { a }{ b })^{ n } } \) = \( \frac { { a }^{ n } }{ { b }^{ n } } \) for every positive integer n.

Example : Evaluate:

(i) \( {(\frac { 3 } { 7 })^{ 3 } } \)    (ii) \( {(\frac { -2 }{ 5 })^{ 3 } } \)

Solution.

(i) \( {(\frac { 3 } { 7 })^{ 3 } } \) = \( \frac { { 3 }^{ 3 } }{ { 7 }^{ 3 } } \) = \(\frac{127}{343}\)
(ii) \( {(\frac { -2 } { 5 }) ^{ 3 } } \) = \( \frac { { (-2) }^{ 3 } }{ { 5 }^{ 3 } } \)
= \( -\frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } } \)
= \(-\frac{8}{125}\)

Negative Integral Exponent of a Rational Number

Let \(\frac{a}{b}\) be any rational number and n be a positive integer.

Then, we define,  \( {(\frac { a }{ b })^{ -n } } \) = \( {(\frac { b }{ a })^{ n } } \)

Example : Evaluate:

(i) \( {(\frac { 1 } { 2 })^{ -3 } } \)    (ii) \( {(\frac { 2 } { 7 })^{ -2 } } \)

Solution.

(i) \( {(\frac { 1 }{ 2 })^{ -3 } } \)
= \( {(\frac { 2 } { 1 })^{ 3 } } \) = \( \frac { { 2}^{ 3 } }{ { 1 }^{ 3 } } \)
= 8
(ii) \( {(\frac { 2 } { 7 })^{ -2 } } \)
= \( {(\frac { 7 } { 2 })^{ 2 } } \)

= \( \frac { { 7}^{ 2 } }{ { 2 }^{ 2 } } \)
= \(\frac{49}{4}\)

Posted in Maths  • 

Vibrations – Important Questions

Q1: Define amplitude.

Ans. It is the maximum displacement of a vibrating body on one side.

Q2: What is called time period?

 

Ans. It is the time taken by a vibrating body to complete one oscillation.

Q3: Define frequency.

Ans. It is the number of vibrations produced by a vibrating body in one second.

Q4: What are ultrasonic waves?

Ans. It is the sound of frequencies more than 20000 hertz. We cannot hear such sounds.

Q5: The maximum displacement of an oscillating object is called the ……… .

Ans. Amplitude.

Q6:The number of oscillations per second made by the object is called the …….…. .

Ans. Frequency.

Q7: What do you mean by the term loudness?

Ans. The sound of high amplitude is called loudness.

Q8: Pendulum A makes 14 oscillations in five seconds and pendulum B makes 10 oscillations in three seconds. Which has a higher frequency? Express the frequency of each pendulum in hertz.

Ans.

Frequency of A = \(\frac{14}{5}\) = 2.8 hertz

Frequency of B = \(\frac{10}{3}\) = 3.33 hertz

Therefore,    Pendulum B has higher frequency.

Q9: Define Hertz.

Ans. Frequency of a vibration or oscillation is measured in hertz. It is the unit of frequency.

Floatation– Higher Order Thinking Skills

Q1. Why does a sharp knife cut better than a blunt knife?

Ans. A sharp knife has a very thin edge to its blade. A sharp knife cuts objects better because due to its very thin edge, the force of our hand falls over very small area of the object producing a large pressure. And this large pressure cuts the object easily. On the other hand, a blunt knife has a thicker edge. A blunt knife does not cut an object easily because due to its thicker edge, the force of our hand falls over a larger area of the object and produces lesser pressure. This
lesser pressure cuts the object with difficulty.

Q2. Why is the tip of the needle sharp?

Ans. The tip of the needle is sharp so that due to its sharp tip, the needle may put the force on a very small area of the cloth, producing a large pressure sufficient enough to pierce the cloth being stitched.

Q3. Why is the tip of the nail pointed?

Ans. A nail has a pointed tip, so that when it is hammered, the force of hammer falls on a very small area of wood creating a large pressure which pushes the nail into wood.

Q4. Why is the pressure on ground more when a man is walking than when he is standing?

Ans. When a man is walking, then at one time only his one foot is on the ground. Due to this, the force of weight of the man falls on a smaller area of the ground and produces more pressure on the ground. On the other hand, when the man is standing both his feet are on the ground and produces lesser pressure on the ground.

Q5. Why is depression much more when a man stands on the cushion than when he lies down on it?

Ans. When a man stands on a cushion then only his two feet are in contact with the cushion. Due to this, the weight of man falls on a small area of cushion producing a large pressure. This large pressure causes a big depression in  cushion. On the other hand, when the same man is lying on the cushion, then his whole body is in contact with the cushion. In this case, the weight of man falls on a much larger area of the cushion producing much smaller pressure. And this smaller pressure produces a very little depression in the cushion.

Q6. Why is it easier to walk on soft sand if we have flat shoes rather than shoes with sharp heels?

Ans. This ¡s because flat shoes have a greater area in contact with the soft sand due to which there is less pressure on the soft ground. Due to this the flat shoes do not sink much in soft sand and it is easy to walk on it. On the other hand, a sharp heel has a small area in contact with the soft sand and so exerts greater pressure on the soft sand. Due to this greater pressure, the sharp heels tend to sink deep into soft sand making it difficult for the wearer to walk on sand.

Q7. Why do tractors have broad tyres?

Ans. The tractors have broad tyres so that there is less pressure on the ground and the tyres do not sink into comparatively soft ground in the fields.

Q8. Why do snow shoes stop you from sinking into snow?

Ans. The snow shoes have large, flat soles so that there is less pressure on the soft snow and stop the wearer from sinking into it.

Q9. Why does a ship made of iron and steel float in water whereas a small piece of iron sinks in it?

Ans. Ship is not a solid block of iron and steel. A ship is a hollow object made of iron and steel which Contains a lot of air in it. Air has a very low density. Due to the presence of a lot of air in it, the average density of the ship becomes less
than the density of water, therefore, a ship floats in water.

Q10. A boy gets into a floating boat.

(a) What happens to the boat?

(b) What happens to the weight of water displaced?

(c) What happens to the buoyant force on the boat?

Ans. (a) The boat floats lower in water.
(b) The weight of water displaced increases.
(c) The buoyant force on boat increases.

Q11. A 0.5 kg sheet of tin sinks in water but if the same sheet is converted into a box, it floats. Why?

 

Ans. The average density of hollow box made up of sheet of tins becomes less than the density of water, so it floats.

Q12. Why does the driver wear metallic cover while going into the sea?

Ans. The pressure increases with increase of depth of water. The increase in pressure is 1 atm by 10 m increase of water. If this pressure falls on the driver directly, he will not tolerate it. Therefore, if a driver wants to go to large
depth, he should wear the metallic cover. The pressure of water column falls on the metallic cover and the pressure inside the cover remains nearly equal to atmospheric pressure. Thus, by wearing metallic cover the driver remains quite safe.

Q13. Why are the walls of a dam made thick at the bottom and thin upwards?

Ans. The pressure at a point inside a liquid depends on the depth of point from the free surface, therefore, the pressure is very high at the bottom of the dam. To tolerate this pressure, the walls of dam are made thick at the bottom. As
we move upwards, the pressure goes on decreasing: so the thickness of wall is made smaller and smaller. That is why the walls of dam are made thick at the bottom and thin at the top.

Q14. Why is it easier to float in sea water than in river water?

Ans. The body of man is lighter than water but the head is heavier than water. Therefore, total weight of man happens to be greater than the weight of water displaced: therefore the man can sink in water. Hence, for floating in water, the man has to displace more water by moving his arms and legs. By doing so when the weight of water displaced by man becomes equal to the weight of man, the man begins to float. The density of sea water is greater than the fresh water of river, therefore the weight of sea water displaced becomes easily equal to the weight of man. Hence, it is easier to float in sea water than in river water.

Friction – Important Questions

Q1: Fill in the blanks:

(a) Friction opposes the ………. between the surfaces in contact with each other.

(b) Friction depends on the ………. of surfaces.

(c) Friction produces ………. .

(d) Sprinkling of powder on the carrom board ………. friction.

(e) Sliding friction is ………. than the static friction.

Ans.

(a) relative motion

(b) smoothness (or irregularities or nature)

(c) heat

(d) reduces

(e) smaller.

Q2: Four children were asked to arrange forces due to rolling, static and sliding frictions in a decreasing order. Their arrangements are given below. Choose the correct arrangement.

(a) rolling, static, sliding

(b) rolling, sliding, static

(c) static, sliding, rolling

(d) sliding, static, rolling

Ans. (c) static, sliding, rolling.

Q3: Alida runs her toy car on dry marble floor, wet marble floor, newspaper and towel spread on the floor. The force of friction acting on the car on different surfaces in increasing order will be

(a) wet marble floor, dry marble floor, newspaper and towel.

(b) newspaper, towel, dry marble floor, wet marble floor.

(c) towel, newspaper, dry marble floor, wet marble floor.

(d) wet marble floor, dry marble floor, towel, newspaper.

Ans. (a) wet marble floor, dry marble floor, newspaper and towel.

Q4: Suppose your writing desk is tilted a little. A book kept on it starts sliding down. Show the direction of frictional force acting on it.

Ans. Frictional force will act upward i.e., direction opposite to that of sliding book.

Q5: Explain why sportsmen use shoes with spikes.

Ans. It is done to provide the shoes better grip on the ground.

Q6: Name the force responsible for wearing out of bicycle tyres.

Ans. Frictional force.

Q7: Answer the following

(a) A ball is rolling on a road towards the north. What will be the direction of the frictional force acting on it?

(b) If the direction of movement in the above case is changed to the south by applying force, then what will be the direction of the frictional force?

Ans.

(a) South

(b) North.

Q8: What is the cause of friction?

Ans. The cause of friction is the irregularities on the two surfaces in contact.

Q9: On which kind of surface the friction will be greater (i) smooth surface (ii) rough surface.

Ans. Friction will be greater in case of a rough surface.

Q10: Does the use of ball bearing reduce or increase friction?

Ans. Use of ball bearing reduces the friction.

Q11: Why do we sprinkle fine powder on the carrom board?

Ans. We want to reduce friction in order to increase efficiency.

Q12: What are lubricants?

Ans. The substances which reduce friction are called lubricants.

Q13: What is drag?

Ans. The frictional force exerted by fluids is also called drag.

Short Cut Method for Finding The Cubes Of a Two- Digit Number

We have :  \( { (a+b) }^{ 3 } \) = \( { a }^{ 3 } \) + 3 \( { a }^{ 2 } \) b+ 3 a \( { b }^{ 2 } \) + \( { b }^{ 3 } \).
Method : For finding the cube of a two-digit number with the tens digit = a and the units digit b, we make four columns, headed by
\( { a }^{ 3 } \), 3 \( { a }^{ 2 } \) b, 3 a \( { b }^{ 2 } \) and \( { b }^{ 3 } \).
The rest of the procedure is the same as followed in squaring a number by the column method.

We simplify the working as :

Short-Cut-Method-For-Finding-The-Cube-Of-A-Two-Digit-Number

Example 1: Find the value of \( { (29) }^{ 3 } \) by the short-cut method.
Solution. Here,            a=2 and b=9

Short-Cut-Method-For-Finding-The-Cube-Of-A-Two-Digit-Number-example-1

Therefore,        \( { (29) }^{ 3 } \) = 24389.

Example 2: Find the value of \( { (71) }^{ 3 } \) by the short-cut method.
Solution. Here, a = 7 and b = 1.

Short-Cut-Method-For-Finding-The-Cube-Of-A-Two-Digit-Number-example-2

Therefore,     \( { (71) }^{ 3 } \) = 357911.

Posted in Maths  • 

Sound – Important Questions

Q1: A pendulum oscillates 40 times in 4 seconds. Find its time period and frequency.

Ans.
    Given that,
Number of oscillation    = 40

Total time taken        = 4 seconds

Time period                = time taken in one oscillation

= \(\frac{Total  time}{Total   number   of  oscillations}\)

= \(\frac{4 seconds}{40}\)

= \(\frac{1}{10}\) = 0.1 second

 

Again, frequency       = number of oscillations per second

= \(\frac{Number  of  vibrations }{ Time  taken}\)

= \(\frac{40}{4}\)

= 10 per second = 10 Hz.

Q2: The sound from a mosquito is produced when it vibrates its wings at an average rate of 500 vibrations per second. What is the time period of the vibration?

Ans.

Number of vibrations per second = 500

Time period = time taken for one vibration

= \(\frac{Total  time}{Total   number   of  vibrations}\)
= \(\frac{1}{500}\) Hz
= 0.002 Hz.

Q3: Identify the part which vibrates to produce sound in the following instruments:

        (a) Dholak            (b) Sitar            (c) Flute

 

Ans.

(a) Stretched membrane

(b) String of sitar

(c) Air column.

Q4: What do you understand by stringed instruments?

Ans. Those instruments have taut strings, which vibrate when they are plucked, struck, or played with a bow.

Q5: Which object is vibrating when the following sounds are produced?

(i) The sound of a sitar or veena.

(ii) The sound of the tabla.

(iii) The sound of a school bell.

(iv) The buzzing of a bee or a mosquito.

(v) The sound of a bursting balloon.

(vi) The radio.

Ans.

(i) Strings of sitar or veena vibrate to produce sound.

(ii) The skin of the tabla vibrate to produce sound.

(iii) The disc of the bell when beated with hammer vibrate to produce sound.

(iv) Wings of the mosquito or a bee vibrate to produce sound.

(v) Air vibrates to produce sound.

(vi) In the radio speakers vibrate to produce sound.

Q6: What is the normal length of vocal cords in man?

Ans. The normal length of the vocal cord of a man is about 20 mm long.

Q7: What is the unit of the loudness of sound? Explain giving the loudness of few sounds.

Ans. The loudness is expressed in a unit called decibel (dB). It is essentially a ratio if the actual loudness to some base loudness level. The base level is usually defined as that loudness of sound that the human ear can just perceive. The following table gives some idea of the loudness of sound coming from various sources:

Loudness-of-sound-unitsAt 80 dB the noise becomes physically painful.

Q8: We hear sound only if its frequency is greater than ……. and lower than ……. .

Ans. 20 hertz, 20000 hertz.

Q9: A tight membrane produces sound of ……. frequency than a loose one.

Ans. Higher.