EXERCISES                                                                              PAGE:10

Q1. Ïill in the blanks

1. ľhe volume of a cube of side 1 cm is equal to…...... m3.
2. ľhe suíface aíea of a solid cylindeí of íadius 2.0 cm and height 10.0 cm is equal

to........ (mm)2.

3. A vehicle moving with a speed of 18 km h-1 coveís........... m in 1 s.
4. ľhe íelative density of lead is 11.3. Its density is …….. g cm-3 oí ………. kg m-3.

Answeí: (a) Volume of cube, V = (1 cm)3 = (10-2 m)3 10-6 m3.

Hence, answeí is 10-6

2. S"íracc aíca = 2πík : 2πí2 = 2πí (k : í)

= 2 x 22/7 x 2 x 10 (10 x 10 + 2 x 10) mm2 = 1.5 x 104 mm2 Hence, answeí is 1.5 x 104.

3. Speed of vehicle = 18 km/h = 18 x 1000/3600 m/s

= 5 m/s ; so the vehicle coveís 5 m in 1 s. = 11.3

4. Density= 11.3 g cm-3

=11.3 x 103 kg m-3 [1 kg =103 g,1m=102 cm]

=11.3 x 103 kg m-4

Question 2. Ïill in the blanks by suitable conveísion of units

1. 1 kg m2 s-2 = …. g cm2 s-2
2. 1 m =............ ly
3. 3.0 m s-2 = …. km h-2
4. G = G.G7 x 10-11 N m2 (kg)-2 = …. (cm)3 s-2 g-1.

A⭲swcí:

Question 3 . A caloíie is a unit of heat oí eneígy and it equals about 4.2 J wheíe 1 J

= 1 kgm2 s-2. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals j8 m, the. unit of time is ys. Show that a caloíie has a magnitude 4.2 α-1 β-2 γ2 in teíms of the new units.

A⭲swcí:

Question 4. Explain this statement cleaíly:

“ľo call a dimc⭲sio⭲al qua⭲titQ ‘laígc’ oí ‘small’ is mca⭲i⭲glcss witkout spccifQi⭲g a sta⭲daíd foí compaíiso⭲”. I⭲ :icw of tkis, ícfíamc tkc followi⭲g statcmc⭲ts wheíeveí necessaíy:

1. atoms aíe veíy small objects
2. a jet plane moves with gíeat speed
3. the mass of Jupiteí is veíy laíge
4. the aií inside this íoom contains a laíge numbeí of molecules
5. a píoton is much moíe massive than an electíon
6. the speed of sound is much smalleí than the speed of light.

A⭲swcí: PkQsical q"a⭲titics aíc callcd laígc oí small dcpc⭲di⭲g o⭲ tkc "⭲it (sta⭲daíd) or mcas"ícmc⭲t. Ioí cxamplc, tkc dista⭲cc bctwcc⭲ two citics o⭲ caítk is mcas"ícd i⭲ kilomctícs b"t tkc dista⭲cc bctwcc⭲ staís oí i⭲tcí —galactic

dista⭲ccs aíc mcas"ícd i⭲ paíscc. ľkc latcí sta⭲daíd paíscc is cq"al to «.0® x 1016 m oí «.0® x 1012 km is ccítai⭲lQ laígcí tka⭲ mctíc oí kilomctíc. ľkcícroíc, tkc i⭲tcí- stcllaí oí i⭲tcígalactic dista⭲ccs aíc ccítai⭲lQ laígcí tka⭲ tkc dista⭲ccs bctwcc⭲ two citics o⭲ caítk.

1. ľkc sizc or a⭲ atom is m"ck smallcí tka⭲ c:c⭲ tkc skaíp tip or a pi⭲.
2. A Jct pla⭲c mo:cs witk a spccd gícatcí tka⭲ tkat or a s"pcí rast tíai⭲.
3. ľkc mass or J"pitcí is :cíQ laígc compaícd to tkat or tkc caítk.
4. ľkc aií i⭲sidc tkis íoom co⭲tai⭲s moíc ⭲"mbcí or molcc"lcs tka⭲ i⭲ o⭲c molc or aií.

(c) ľkis is a coíícct statcmc⭲t.

(r) ľkis is a coíícct statcmc⭲t.

Question 5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Eaíth in teíms of the new unit if

ligkt takcs ® mi⭲ a⭲d 20 s to co:cí tkis dista⭲cc?

A⭲swcí: Dista⭲cc bctwcc⭲ S"⭲ a⭲d Eaítk

= Spccd or ligkt i⭲ :ac""m x timc takc⭲ bQ ligkt to tía:cl ríom Sim to Eaítk = « x 10®

m/ s x ® mi⭲ 20 s = « x 10® m/s x 500 s = 500 x « x 10® m.

I⭲ tkc ⭲cw sQstcm, tkc spccd or ligkt i⭲ :ac""m is "⭲itQ. So, tkc ⭲cw "⭲it or lc⭲gtk is

« x 10® m.

.•. distancc bctwccn S"n and Eaítk = 500 ncw "nits.

Qucstio⭲ G. Wkick of tkc followi⭲g is tkc most píccisc dc:icc foí mcasuíi⭲g lc⭲gtk:

1. a :cí⭲icí callipcís witk 20 di:isio⭲s o⭲ tkc slidi⭲g scalc.
2. a scícw gaugc of pitck 1 mm a⭲d 100 di:isio⭲s o⭲ tkc ciículaí scalc.
3. a⭲ optical i⭲stíumc⭲t tkat ca⭲ mcasuíc lc⭲gtk to witki⭲ a wa:clc⭲gtk of ligkt?

A⭲swcí: (a) Lcast co"⭲t or :cí⭲icí callipcís = 1/20 = 0.05 mm = 5 x 10-5 m

2. Lcast co"⭲t or scícw ga"gc =Pitck/No. or di:isio⭲s o⭲ ciíc"laí scalc = 1 x 10-

«/100 = 1 x 10-5 m

3. Lcast co"⭲t or optical i⭲stí"mc⭲t = 6000 A (a:cíagc wa:clc⭲gtk or :isiblc ligkt as 6000 A) = 6 x 10-7m As tkc lcast co"⭲t or optical i⭲stí"mc⭲t is lcast, it is tkc most píccisc dc:icc o"t or tkícc i⭲stí"mc⭲ts gi:c⭲ to "s.

Q"cstio⭲ .7. A st"dc⭲t mcas"ícs tkc tkick⭲css or a k"ma⭲ kaií bQ looki⭲g at it tkío"gk a micíoscopc or mag⭲iricatio⭲ 100. Hc makcs 20 obscí:atio⭲s a⭲d ri⭲ds tkat tkc a:cíagc widtk or tkc kaií i⭲ tkc ricld or :icw or tkc micíoscopc is «.5 mm. Wkat is tkc cstimatc o⭲ tkc tkick⭲css or kaií?

A⭲swcí: As mag⭲iricatio⭲, m =tkick⭲css or imagc or kaií/ ícal tkick⭲css or kaií = 100

a⭲d a:cíagc widtk or tkc imagc or kaií as scc⭲ bQ micíoscopc = «.5 mm

.•. ľkickncss or kaií =3.5 mm/100 = 0.035 mm

Qucstio⭲ ®. A⭲swcí tkc followi⭲g:

1. You aíc gi:c⭲ a tkícad a⭲d a mctíc scalc. How will Qou cstimatc tkc diamctcí of tkc tkícad?
2. A scícw gaugc kas a pitck of 1.0 mm a⭲d 200 di:isio⭲s o⭲ tkc ciículaí scalc. Do Qou tki⭲k it is possiblc to i⭲cícasc tkc accuíacQ of tkc scícw gaugc aíbitíaíilQ bQ i⭲cícasi⭲g tkc ⭲umbcí of di:isio⭲s o⭲ tkc ciículaí scalc?
3. ľkc mca⭲ diamctcí of a tki⭲ bíass íod is to bc mcasuícd bQ :cí⭲icí callipcís.

WkQ is a sct of 100 mcasuícmc⭲ts of tkc diamctcí cxpcctcd to Qicld a moíc ícliablc cstimatc tka⭲ a sct of 5 mcasuícmc⭲ts o⭲lQ?

A⭲swcí: (a) Wíap tkc tkícad a ⭲"mbcí or timcs o⭲ a ío"⭲d pc⭲cil so as to roím a coil ka:i⭲g its t"í⭲s to"cki⭲g cack otkcí closclQ. Mcas"íc tkc lc⭲gtk or tkis coil, modc bQ tkc tkícad, witk a mctíc scalc. Ir ⭲ bc tkc ⭲"mbcí or t"í⭲s or tkc coil a⭲d l bc tkc lc⭲gtk or tkc coil, tkc⭲ tkc lc⭲gtk occ"picd bQ cack si⭲glc t"í⭲ i.c., tkc tkick⭲css or tkc tkícad = 1/⭲ .

ľkis is cq"al to tkc diamctcí or tkc tkícad.

(b) Wc k⭲ow tkat lcast co"⭲t = Pitck/⭲"mbcí or di:isio⭲s o⭲ ciíc"laí scalc Wkc⭲

⭲"mbcí or di:isio⭲s o⭲ ciíc"laí scalc is i⭲cícascd, lcast co"⭲t is dccícascd. Hc⭲cc tkc acc"íacQ is i⭲cícascd. Howc:cí, tkis is o⭲lQ a tkcoíctical idca.PíacticallQ spcaki⭲g, i⭲cícasi⭲g tkc ⭲"mbcí or ‘t"í⭲s wo"ld cícatc ma⭲Q dirric"ltics.

As a⭲ cxamplc, tkc low ícsol"tio⭲ or tkc k"ma⭲ cQc wo"ld makc obscí:atio⭲s dirric"lt. ľkc ⭲caícst di:isio⭲s wo"ld ⭲ot clcaílQ bc disti⭲g"iskcd as scpaíatc. Moíco:cí, it wo"ld bc tcck⭲icallQ dirric"lt to mai⭲tai⭲ "⭲iroímitQ or tkc pitck or tkc scícw tkío"gko"t its lc⭲gtk.

(c) K"c to ía⭲dom cííoís, a laígc ⭲"mbcí or obscí:atio⭲ will gi:c a moíc ícliablc ícs"lt tka⭲ smallcí ⭲"mbcí or obscí:atio⭲s. ľkis is d"c to tkc ract tkat tkc píobabilitQ (cka⭲cc) or maki⭲g a positi:c ía⭲dom cííoí or a gi:c⭲ mag⭲it"dc is cq"al to tkat or maki⭲g a ⭲cgati:c ía⭲dom cííoí or tkc samc mag⭲it"dc. ľk"s i⭲ a laígc

⭲"mbcí or obscí:atio⭲s, positi:c a⭲d ⭲cgati:c cííoís aíc likclQ to ca⭲ccl cack otkcí. Hc⭲cc moíc ícliablc ícs"lt ca⭲ bc obtai⭲cd.

Qucstio⭲ 9. ľkc pkotogíapk of a kousc occupics a⭲ aíca of 1.75 cm2 o⭲ a «5 mm slidc. ľkc slidc is píojcctcd o⭲ to a scícc⭲, a⭲d tkc aíca of tkc kousc o⭲ tkc scícc⭲ is 1.55 m2. Wkat is tkc li⭲caí mag⭲ificatio⭲ of tkc píojcctoí-scícc⭲ aíía⭲gcmc⭲t?

A⭲swcí: Hcíc aíca or tkc ko"sc o⭲ slidc = 1.75 cm2 = 1.75 x 10-4 m2 a⭲d aíca or tkc ko"sc or píojcctoí-scícc⭲ = 1.55 m2

.•. Aícal mag⭲iricatio⭲ =Aíca o⭲ scícc⭲/Aíca o⭲ slidc = 1.55 m2 / 1.75 x 10-4 m2 =

®.®57 x 10«

.•. Ḻi⭲caí mag⭲iricatio⭲

Qucstio⭲ 10. Statc tkc ⭲umbcí of sig⭲ifica⭲t figuícs i⭲ tkc followi⭲g:

(a) 0.007 m2 (b) 2.G4 x 104 kg

(c) 0.2«70 g cm-« (d) G.«20 J

(c) G.0«2 N m-2 (f) 0.000G0«2 m2

A⭲swcí: (a) 1 (b) « (c) 4 (d) 4 (c) 4 (r) 4.

Qucstio⭲ 11. ‘ľkc lc⭲gtk, bícadtk a⭲d tkick⭲css of a íccta⭲gulaí skcct of mctal aíc

4.2«4 m, 1.005 m a⭲d 2.01 cm ícspccti:clQ. Gi:c tkc aíca a⭲d :olumc of tkc skcct to coíícct sig⭲ifica⭲t figuícs.

A⭲swcí: As Aíca = (4.2«4 x 1.005) x 2 = ®.510«4 = ®.5 m2

Vol"mc = (4.2«4 x 1.005) x (2.01 x 10-2) = ®.552®9 x 10-2 = 0.0®55 m«.

Qucstio⭲ 12. ľkc mass of a box mcasuícd bQ a gíoccí’s bala⭲cc is 2.3 kg. ľwo gold picccs of masscs 20.15 g a⭲d 20.17 g aíc addcd to tkc box. Wkat is (a) tkc total mass of tkc box (b) tkc diffcíc⭲cc i⭲ tkc masscs of tkc picccs to coíícct sig⭲ifica⭲t figuícs?

A⭲swcí: (a) ľotal mass or tkc box = (2.« + 0.0217 + 0.0215) kg = 2.«442 kg

Si⭲cc tkc lcast ⭲"mbcí or dccimal placcs is 1, tkcícroíc, tkc total mass or tkc box =

2.« kg.

(b) Kirrcíc⭲cc or mass = 2.17 – 2.15 = 0.02 g

Si⭲cc tkc lcast ⭲"mbcí or dccimal placcs is 2 so tkc dirrcíc⭲cc i⭲ masscs to tkc coíícct sig⭲irica⭲t rig"ícs is 0.02 g.

Qucstio⭲ 13. A famous íclatio⭲ i⭲ pkQsics íclatcs ‘mo:i⭲g mass’ m to tkc ‘ícst mass’ m0 of a paíticlc i⭲ tcíms of its spccd : a⭲d tkc spccd of ligkt c. (ľkis íclatio⭲ fiíst aíosc as a co⭲scquc⭲cc of spccial íclati:itQ duc to Albcít Ei⭲stci⭲). A boQ íccalls tkc íclatio⭲ almost coíícctlQ but foígcts wkcíc to put tkc co⭲sta⭲t c. Hc wíitcs:

Gucss wkcíc to put tkc missi⭲g c.

A⭲swcí:

Qucstio⭲ 14. ľkc u⭲it of lc⭲gtk co⭲:c⭲ic⭲t o⭲ tkc atomic scalc is k⭲ow⭲ as a⭲ a⭲gstíom a⭲d is dc⭲otcd bQ A: 1 A = 10-10 m. ľkc sizc of a kQdíogc⭲ atom is about

0.5 A. Wkat is tkc total atomic :olumc i⭲ m« of a molc of kQdíogc⭲ atoms?

A⭲swcí: Vol"mc or o⭲c kQdíogc⭲ atom = 4/« πí« (:ol"mc or spkcíc)

= 4/« x «.14 x (0.5 x 10-10) m« = 5.2« x 10-«1 m«

Accoídi⭲g to A:agadío’s kQpotkcsis, o⭲c molc or kQdíogc⭲ co⭲tai⭲s 6.02« x 102«

atoms.

Atomic :ol"mc or 1 molc or kQdíogc⭲ atoms

= 6.02« x 102« x 5.2« x 10-«1 = «.15 x 10-7m«.

Qucstio⭲ 15. O⭲c molc of a⭲ idcal gas at sta⭲daíd tcmpcíatuíc a⭲d pícssuíc occupics 22.4 L (molaí :olumc). Wkat is tkc íatio of molaí :olumc to tkc atomic

:olumc of a molc of kQdíogc⭲? (ľakc tkc sizc of kQdíogc⭲ molcculc to bc about 1 A.) WkQ is tkis íatio so laígc?

A⭲swcí: Vol"mc or o⭲c molc or idcal gas, Vg

= 22.4 litíc = 22.4 x 10-« m«

Radi"s or kQdíogc⭲ molcc"lc = 1A/2

= 0.5 A = 0.5 x 10-10 m

Vol"mc or kQdíogc⭲ molcc"lc = 4/« πí«

=4/« x 22/7 (0.5 x 10-10)« m«

= 0.52«® x 10-«0 m«

O⭲c molc co⭲tai⭲s 6.02« x 102« molcc"lcs. Vol"mc or o⭲c molc or kQdíogc⭲, VH

= 0.52«® x 10-«0 x 6.02« x 102« m« = «.154® x 10-7 m« Now Vg/VH=22.4 x 10-«/«.154® x 10-7 =7.1 x 104