Q: Question 17The S.I. unit of pressure is:Kg m-2s-2Kg m-1s-2Kg m2Kg ms-2

Kg m-1s-2Reason — Pressure = forcearea\dfrac{\text{force}}{\text{area}}areaforce = Kg m-1s-2

Question 1

Question 17The S.I. unit of pressure is:Kg m-2s-2Kg m-1s-2Kg m2Kg ms-2

Accepted Answer

Kg m-1s-2Reason — Pressure = forcearea\dfrac{	ext{force}}{	ext{area}}areaforce​ = Kg m-1s-2

Question 2

Question 16Complete the following —1 nano second = ________s1 µs = _______s1 mean solar day = ________s1 year = ________s

Accepted Answer

1 nano second = 10-9 s1 µs = 10-6 s1 mean solar day = 86400 s1 year = 3.15 x 107 s

Question 3

Question 18Write the derived units of the following —SpeedForceWorkPressure

Accepted Answer

The derived units of the following quantities are as follows —QuanityDerived unitSpeedms-1Forcekg m s-2Workkg m2s-2Pressurekg m-1s-2

Question 4

Question 19How are the following derived units related to the fundamental units?NewtonWattJoulePascal

Accepted Answer

Derived unitFundamental unitNewtonkg m s-2Wattkg m2s-3Joulekg m2s-2Pascalkg m-1s-2

Question 5

Question 20Name the physical quantities related to the following units —km2newtonjoulepascalwatt

Accepted Answer

The physical quantities related to the following units are —UnitPhysical quantitykm2areanewtonforcejouleenergypascalpressurewattpower

Question 6

Question 2A vernier has 10 divisions and they are equal to 9 divisions of main scale in length. If the main scale is calibrated in mm, what is its least count?

Accepted Answer

(i) As we know,L.C.=Value of 1 main scale div (x)Total no. of div on vernier (n)\text{L.C.} = \dfrac{\text{Value of 1 main scale div (x)}}{\text{Total no. of div on vernier (n)}}L.C.=Total no. of div on vernier (n)Value of 1 main scale div (x)​Given,Total number of divisions on vernier = 10Value of one main scale division(x) = 1 mmSubstituting the values in the formula given above we get,L.C.=1mm10⇒L.C.=0.1mm⇒L.C.=0.01cm\text{L.C.} = \dfrac{\text{1mm}}{\text{10}} \[0.5em] \Rightarrow \text{L.C.} = 0.1 \text{mm} \[0.5em] \Rightarrow \text{L.C.} = 0.01 \text{cm} \[0.5em]L.C.=101mm​⇒L.C.=0.1mm⇒L.C.=0.01cmHence, least count of the vernier callipers is 0.01cm.

Question 7

Question 10The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.(i) What is the pitch of the screw gauge?(ii) What is the least count of the screw gauge?

Accepted Answer

(i) As we know,Pitch = distance moved ahead in 1 revolutionGiven,Distance covered in two revolutions = 1 mm∴, we get,Pitch=12 mm=0.5 mm\text {Pitch} = \dfrac{1}{2} \text{ mm} = 0.5 \text{ mm} \[0.5em]Pitch=21​ mm=0.5 mmHence, pitch of the screw gauge = 0.5 mm(ii) As we know,Least Count=PitchTotal no. of div on circular head\text{Least Count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\[0.5em]Least Count=Total no. of div on circular headPitch​Given,Pitch = 0.5 mmTotal number of divisions on circular scale = 50Substituting the values in the formula above we get,Least count=0.550Least count=0.01 mm\text {Least count} = \dfrac{0.5}{50} \[0.5em] \text {Least count} = 0.01 \text { mm} \[0.5em]Least count=500.5​Least count=0.01 mmHence, the least count of the screw gauge = 0.01 mm

Question 8

Question 11The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with the base line.Find —The least count, andThe diameter of the wire

Accepted Answer

(i) As we know,Least Count=PitchTotal no. of div on circular head\text{Least Count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\[0.5em]Least Count=Total no. of div on circular headPitch​Given,Pitch = 1 mmNumber of divisions on circular head = 100Substituting the values in the formula above we get,Least count=1100mmLeast count=0.01mm=0.001cm\text {Least count} = \dfrac{1}{100} \text{mm} \[0.5em] \text {Least count} = 0.01 \text {mm} = 0.001 \text {cm}\[0.5em]Least count=1001​mmLeast count=0.01mm=0.001cmHence, the least count of the screw gauge = 0.001 cm(ii) As we know,Diameter of the wire = main scale reading + circular scale reading     [Equation 1]andCircular scale reading = p x L.C.     [Equation 2]p = 45L.C. = 0.001 cmSubstituting the values in the Equation 2 we get,Circular scale reading = 45 x 0.001 = 0.045Hence, circular scale reading = 0.045 cmGiven, main scale reads 2 mm = 0.2 cmHence, main scale reading = 0.2 cmSubstituting the values in Equation 1 we get,Diameter of the wire = 0.2 cm + 0.045 cm = 0.245 cmHence, the diameter of the wire is 0.245 cm.

Question 9

Question 12When a screw gauge of least count 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions.What is the diameter of the wire in cm?If the zero error is +0.005 cm, what is the correct diameter?

Accepted Answer

As we know,Diameter of the wire = main scale reading + circular scale reading     [Equation 1]andReading on thimble = p x L.C.     [Equation 2]Given,p = 27L.C. = 0.01 mm = 0.001 cmSubstituting the values in the Equation 2 we get,reading on thimble = 27 x 0.001 = 0.027 cmHence, reading on thimble = 0.027 cm and reading on sleeve = 1 mm = 0.1 cmUsing Equation 1 we get,Diameter of the wire = 0.1 cm + 0.027 cm = 0.127 cmHence, the diameter of the wire is 0.127 cm.(ii) As we know,Correct reading = observed reading - zero errorGiven,Zero error = + 0.005 cmSubstituting the value of zero error in the formula above we get,Correct reading = 0.127 cm - 0.005 cm = 0.122 cmHence, the correct diameter of the wire is 0.122 cm

Question 10

Question 13A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two rotations. When the flat end of the screw is in contact with the stud, the zero of circular scale lies below the base line and 4th division of circular scale is in line with the base line.Find —(i) the pitch,(ii) the least count and(iii) the zero error, of the screw gauge.

Accepted Answer

As we know,Pitch = Distance moved ahead in 1 revolution.Given,Total number of divisions on circular scale = 50Distance covered in two revolutions = 1 mm∴, we get,Pitch=12 mm=0.5 mm\text {Pitch} = \dfrac{1}{2} \text{ mm} = 0.5 \text{ mm} \[0.5em]Pitch=21​ mm=0.5 mmHence, pitch of the screw gauge = 0.5 mm(ii) As we know,Least count=PitchTotal no. of div on circular head\text{Least count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\[0.5em]Least count=Total no. of div on circular headPitch​Here,Pitch = 0.5 mmNumber of divisions on circular head = 50Substituting the values in the formula above we get,Least count=0.550Least count=0.01mm\text {Least count} = \dfrac{0.5}{50} \[0.5em] \text {Least count} = 0.01 \text {mm} \[0.5em]Least count=500.5​Least count=0.01mmHence, the least count of the screw gauge = 0.01 mm(iii) As we know,Zero error = Coinciding Division x Least CountCoinciding Division = 4L.C. = 0.01 mmSubstituting the values in the formula above we get,Zero error = 4 x 0.01 = + 0.04 mmHence, zero error, of the screw gauge = + 0.04 mm

Question 11

Question 14Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once.Find —(i) pitch of the screw gauge,(ii) least count of the screw gauge, and(iii) the diameter of the wire.

Accepted Answer

(i) As we know,Pitch = distance moved ahead in 1 revolutionand given,Distance covered in one revolutions = 1 mmHence, Pitch of the screw gauge = 1 mm(ii) As we know,Least count=PitchTotal no. of div on circular head\text{Least count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\[0.5em]Least count=Total no. of div on circular headPitch​Here,Pitch = 1 mmNumber of divisions on circular head = 50Substituting the values in the formula above we get,Least count=150Least count=0.02 mm\text {Least count} = \dfrac{1}{50} \[0.5em] \text {Least count} = 0.02 \text { mm} \[0.5em]Least count=501​Least count=0.02 mmHence, the least count of the screw gauge = 0.02 mm(iii) As we know,Diameter of the wire = main scale reading + circular scale reading     [Equation 1]andCircular scale reading = p x L.C.     [Equation 2]p = 47L.C. = 0.02 mmSubstituting the values in the Equation 2 we get,Circular scale reading = 47 x 0.02 = 0.94 mmHence, circular scale reading = 0.94 mm and main scale reading = 4 mm.Using Equation 1 we get,Diameter of the wire = 4 mm + 0.94 mm = 4.94 mmHence, the diameter of the wire is 4.94 mm.

Question 12

Question 15A screw has a pitch equal to 0.5 mm. What should be the number of division on its head so as to read correct up to 0.001mm with its help?

Accepted Answer

As we know,Least count=PitchTotal no. of div on circular head\text {Least count} = \dfrac{\text{Pitch}}{\text{Total no. of div on circular head}} \[0.5em]Least count=Total no. of div on circular headPitch​Given,Pitch = 0.5 mmL.C. = 0.001 mmSubstituting the values in the formula above we get,0.001mm=0.5Total no. of div on circular head⇒Total no. of div on circular head=0.50.001⇒Total no. of div on circular head=5000.001\text{mm} = \dfrac{0.5}{\text{Total no. of div on circular head}} \[0.5em] \Rightarrow \text{Total no. of div on circular head} = \dfrac{0.5}{0.001} \[0.5em] \Rightarrow \text{Total no. of div on circular head} = 5000.001mm=Total no. of div on circular head0.5​⇒Total no. of div on circular head=0.0010.5​⇒Total no. of div on circular head=500Hence, total number of divisions on circular head = 500.

Question 13

Question 1How does the time period (T) of a simple pendulum depend on its length (l) ? Draw a graph showing the variation of T2 with l. How will you use this graph to determine the value of g (acceleration due to gravity)?

Accepted Answer

Time period of a simple pendulum is directly proportional to the square root of its effective length.i.e., T ∝ l\sqrt{	ext l}l​Graph showing the variation of T2 with l is given below:In order to find the acceleration due to gravity with the help of the above graph, we follow the following steps —The slope of the straight line obtained in the T2 vs l graph, as shown in fig, can be obtained by taking two points P and Q on the straight line and drawing normals from these points on the X and Y axes. Then, note the value of T2, say T12 and T22 at a and b respectively, and also the value of l say l1 and l2 respectively at c and d.Then,Slope=PRQR=abcd=T12−T22l1−l2	ext{Slope} = \dfrac{	ext {PR}}{	ext {QR}} = \dfrac{	ext {ab}}{	ext {cd}} = \dfrac{	ext T_1^2 - 	ext T_2^2}{	ext l_1 - 	ext l_2}Slope=QRPR​=cdab​=l1​−l2​T12​−T22​​This slope is found to be a constant at a place and,Slope=4π2g	ext{Slope} = \dfrac{4	ext π^2}{	ext g}Slope=g4π2​where, g = acceleration due to gravity at that place.Thus, g can be determined at a place from the graph using the following relation,g=4π2Slope of T2 vs l graph	ext g = \dfrac{4	ext π^2}{	ext{Slope of } 	ext T^2 	ext{ vs l graph}}g=Slope of T2 vs l graph4π2​

Question 14

Question 1(viii)The length of a simple pendulum is made one-fourth. Its time period becomes —four timesone-fourthdoublehalf

Accepted Answer

halfReason — Time period of a simple pendulum is given by:T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em]T=2πgl​​Time period is directly proportional to the square root of the length of the pendulum.In the case when length is made one-fourth, we see that —T=2πl4×gT=22×πlgT=πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{4\times \text g}} \[0.5em] \text T = \dfrac{2}{2} \times \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em] \text T = \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em]T=2π4×gl​​T=22​×πgl​​T=πgl​​Hence, we can say that when the length is made four times, time period of a simple pendulum is reduced to half.

Question 15

Question 3A seconds' pendulum is taken to a place where acceleration due to gravity falls to one forth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

Accepted Answer

As we know,T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}T=2πgl​​We observe that time period is inversely proportional to the square root of acceleration due to gravity.Hence, when 'g' falls to one-fourth, time period increases.When acceleration due to gravity is reduced to one fourth, we see that —T=2πlg4T=2π4lgT=2×2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\dfrac{\text g}{4}}} \[0.5em] \text T = 2 π \sqrt{\dfrac{4\text l}{\text g}} \[0.5em] \text T = 2 \times 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em]T=2π4g​l​​T=2πg4l​​T=2×2πgl​​Hence, we can say that when acceleration due to gravity is reduced to one fourth, time period of a simple pendulum doubles.As, the given pendulum is a seconds' pendulum so T = 2s∴ New T = 2 x 2 = 4s

Question 16

Question 5Compare the time periods of two pendulums of length 1m and 9m.

Accepted Answer

As we know that,T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em]T=2πgl​​Time period is directly proportional to the square root of the length of the pendulum.In the case when length is 1m,T1=2π1g\text T_1 = 2 \text π \sqrt{\dfrac{1}{\text g}} \[0.5em]T1​=2πg1​​andIn the case when length is 9m,T2=2π9g\text T_2 = 2 \text π \sqrt{\dfrac{9}{\text g}} \[0.5em]T2​=2πg9​​So, comparison of T1 and T2 gives —T1:T2=2π1g:2π9gT1:T2=19⇒T1:T2=13\text T_1 : \text T_2 = 2 \text π \sqrt{\dfrac{1}{\text g}} : 2 \text π \sqrt{\dfrac{9}{\text g}} \[0.5em] \text T_1 : \text T_2 = \sqrt{\dfrac{1}{9}} \[0.5em] \Rightarrow \text T_1 : \text T_2 = \dfrac{1}{3} \[0.5em]T1​:T2​=2πg1​​:2πg9​​T1​:T2​=91​​⇒T1​:T2​=31​Hence, T1 : T2 = 1 : 3

Question 17

Question 6A pendulum completes 2 oscillations in 5s.(a) What is its time period?(b) If g = 9.8 ms-2, find its length.

Accepted Answer

Given,2 oscillations in 5 seconds, sofrequency per second=25⇒frequency per second=0.4 hertz\text {frequency per second} = \dfrac{2}{5} \[0.5em] \Rightarrow \text {frequency per second} = \text {0.4 hertz} \[0.5em]frequency per second=52​⇒frequency per second=0.4 hertzHence, frequency of oscillation = 0.4 hertz.As we know that,T=1f\text T = \dfrac{1}{\text f}T=f1​So, substituting the value of f = 0.4 hertz, in equation above we get,T=10.4⇒T=2.5 s\text T = \dfrac{1}{0.4} \[0.5em] \Rightarrow \text T = 2.5\ \text s \[0.5em]T=0.41​⇒T=2.5 sHence, time period of pendulum is 2.5 s(b) As we know,T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em]T=2πgl​​Given,g = 9.8 ms-2and we know,π = 3.14T = 2.5 sSubstituting the values in the formula above we get,2.5=2×3.14l9.8⇒(2.52×3.14)=l9.8⇒(2.52×3.14)2=l9.8⇒(2.56.28)2=l9.8⇒(0.398)2=l9.8⇒0.158=l9.8⇒l=9.8×0.158⇒l=1.55 m2.5 = 2 \times 3.14 \sqrt{\dfrac{\text l}{9.8}} \[0.5em] \Rightarrow (\dfrac{2.5}{2 \times 3.14}) = \sqrt{\dfrac{\text l}{9.8}} \[0.5em] \Rightarrow (\dfrac{2.5}{2 \times 3.14})^2 = \dfrac{\text l}{9.8} \[0.5em] \Rightarrow (\dfrac{2.5}{6.28})^2 = \dfrac{\text l}{9.8} \[0.5em] \Rightarrow (0.398)^2 = \dfrac{\text l}{9.8} \[0.5em] \Rightarrow 0.158 = \dfrac{\text l}{9.8} \[0.5em] \Rightarrow \text l = 9.8 \times 0.158 \[0.5em] \Rightarrow \text l = 1.55\ \text m2.5=2×3.149.8l​​⇒(2×3.142.5​)=9.8l​​⇒(2×3.142.5​)2=9.8l​⇒(6.282.5​)2=9.8l​⇒(0.398)2=9.8l​⇒0.158=9.8l​⇒l=9.8×0.158⇒l=1.55 mHence, length of a seconds’ pendulum = 1.55 m

Question 18

Question 7The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the ratio of their lengths ?

Accepted Answer

As we know that,T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em]T=2πgl​​Time period is directly proportional to the square root of the length of the pendulum.In the case when,T1 : T2 = 2 : 1     [Equation 1]we know that,T1 : T2 = l1:l2\sqrt{\text l_1} : \sqrt{\text l_2}l1​​:l2​​     [Equation 2]So we get,l1:l2=2:1⇒l1l2=21\sqrt{\text l_1} : \sqrt{\text l_2} = 2 : 1 \[0.5em] \Rightarrow \dfrac{\sqrt{\text l_1}}{\sqrt{\text l_2}} = \dfrac{2}{1}l1​​:l2​​=2:1⇒l2​​l1​​​=12​Squaring both sides we get,l1l2=2212⇒l1l2=41⇒l1:l2=4:1\dfrac{\text l_1}{\text l_2} = \dfrac{2^2}{1^2} \[0.5em] \Rightarrow \dfrac{\text l_1}{\text l_2} = \dfrac{4}{1} \[0.5em] \Rightarrow \text l_1 : \text l_2 = 4 : 1l2​l1​​=1222​⇒l2​l1​​=14​⇒l1​:l2​=4:1Hence, ratio of lengths = 4 : 1

Question 19

Question 9How much time does the bob of a seconds' pendulum take to move from one extreme of its oscillation to the other extreme?

Accepted Answer

As we know that, the time period of a seconds' pendulum is 2s.So, time taken for a seconds’ pendulum to move from one extreme to other is equal to the half of time period.Thalf=T2⇒Thalf=22=1 s\text T_{\text {half}} = \dfrac{\text T}{2} \[0.5em] \Rightarrow \text T_{\text {half}} = \dfrac{2}{2} = 1\ \text s \[0.5em]Thalf​=2T​⇒Thalf​=22​=1 sHence, time taken for a seconds' pendulum to move from one extreme to other extreme = 1s

Question 20

Question 4How is the time period of a simple pendulum affected, if at all, in the following situations:(a) The length is made four times,(b) The acceleration due to gravity is reduced to one - fourth.

Accepted Answer

As we know that,T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}T=2πgl​​where,T = Time periodl = effective length of the pendulumg = acceleration due to gravity(a) In the case when length is made four times, let time period be T1, we see that —T1=2π4lgT1=2×2πlgT1=2×T\text T_1 = 2 \text π \sqrt{\dfrac{4\text l}{\text g}} \[0.5em] \text T_1= 2 \times 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em] \text T_1 = 2 \times \text T \[0.5em]T1​=2πg4l​​T1​=2×2πgl​​T1​=2×THence, we can say that when the length is made four times, time period of a simple pendulum is doubled.(b) In the case, when acceleration due to gravity is reduced to one fourth, let time period be T1, we see that —T1=2πlg4T1=2π4lgT1=2×2πlgT1=2×T\text T_1 = 2 \text π \sqrt{\dfrac{\text l}{\dfrac{\text g}{4}}} \[0.5em] \text T_1 = 2 \text π \sqrt{\dfrac{4\text l}{\text g}} \[0.5em] \text T_1 = 2 \times 2 \text π \sqrt{\dfrac{\text l}{\text g}} \[0.5em] \text T_1 = 2 \times \text T \[0.5em]T1​=2π4g​l​​T1​=2πg4l​​T1​=2×2πgl​​T1​=2×THence, we can say that when acceleration due to gravity is reduced to one fourth, time period of a simple pendulum is doubled.

Measurements and Experimentation

Exercise 1A Multiple Choice Type

Exercise 1A Very Short Answer Type

Exercise 1B Numericals

Exercise 1C Long Answer Type

Exercise 1C Multiple Choice Type

Exercise 1C Numericals

Exercise 1C Very Short Answer Type