00:00अगर आप होगा, क्योंकि आप दिख रहे हैं।
00:03हैलो स्टुड्रेंट से आगर वालोम मी और कोसी सहीचाच शुरुप से बहुत ब्राष्ट होकी हैं।
00:05ऐसा के साथ मैं एक अलग प्लेलिस्ट के आलक प्रकरण कर रहा हूँ।
00:09In this series, I am gonna start current electricity installed NCRT,
00:15पहला वीडियो होगा, फिर तो जकता खेल खेल खेल के लिए ही,
00:19Of course, it is for class 12 CDSC, NCRT exercises हमारा नेक्स वीडियो होने बाला होगा.
00:26आए चलते हैं, इंस्टेल्ट इंजांबल्स को आप अच्छे तरीके से आप मेंज कीजिए।
00:30फिर उसके आद कोई डाउट हो तो आप अगर कुछ विश्वाद नहीं भी होगी तो अगर एकसर्साइज़ करते करते आप काफी ज़्यादा खुले होगे और फिर यह एकसर्साइज़ और एकसर्साइज़ दोनों के लिए आप जी को भी ध्यान में रखना है तो बहुत नह
01:00होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी नहीं होगी न
01:30हमें एन पहले मालूम करना होगा बिकोस ड्रिफ्ट स्पीड से रिलेटेट या तो
01:36हमें आई इस इकल तो एन एड़ी वीडी और वीडी इस माइनस इटाउ पाई एम
01:46हमारे लिए इस उप्शन इस गलोस और कौन-कौन से चीज गिवन हैं यहाँ पर डेंसिटी
01:58कौपर के लिए कौपर के लिए 9 के लिए 3 केजी पर यूबिक मीटर और इस एटाउमिक मास
02:08के लिए 63.5 यू यू यह एटाउमिक मास यूनिट है तो ड्रिफ्ट स्पीड से रिलेटेट या पहले मालूम करना होगा
02:20आप हमें इस से ही ये वर्क आउट करना पड़ेगा टेंड रेंज नम्बर डेंसिटी
02:25या अगर विशा बहुत लिए सीधा पुछना है नम्बर डेंसिटी मुकुल पर लिए बादकाम नम्बर पर लिए बादकाम
02:33पर यूनिट वालूम यहाँ पर मॉल कौन्सेप्ट इंबॉल हो रहा है और नमबर पर यूनिट वालूम
02:46अब नमबर के बारे में कुछ भी बात ही नहीं की गई है कुछ लेकिन उस से रिलेटेट एक चीज़ दे दी गई
02:53पर यूनिट वालूम यहाँ पर मॉल कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन
03:23कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और नमबर के बारे में कौन्सेप्ट इंबॉल हो रहा है और न
03:53मैं उनकिंगी का मात समय किसंगें पर तो हमको मात थो मालूं कर रही सकते था वन एटाम इपट का
04:02ఆ డ్ర్ థా ఆనంది లో లో పోలీ జూజ లో టగొన 2 కొ బ్రా అనూడ్ పున్న థ�ూండ్ పో టే కోలంే ధినెను ప్రవా కౌన్ ఇంటర్ ఇట్ట్ న�
04:32एक परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद प�
05:02परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद परवाद
05:32गराम का मतलब
05:3463.5
05:36upon 1000
05:386 x 10
05:4023 x 9
05:4210 x 6
05:4463.5
05:4663.5
05:48नाव अफ़्टरवर्स
05:50आहव टू कैल्कुलेट
05:52आहव टू कैल्कुलेट
05:5429
05:569 x 6
05:5854 x
06:0063.5
06:028 x 5
06:040
06:0654
06:0863.5
06:100.8
06:125
06:140 x
06:1610 x
06:1829
06:208.5 x
06:2210 x
06:2428
06:26आव अगर वापस से
06:28हम बात करें
06:30from this equation
06:32Vd is equal to I upon N
06:34EA
06:36Vd is equal to
06:38directly
06:40I is 1.5 ampere
06:42N is
06:448.5 x
06:4610 x 28
06:48N is 1.6 x
06:5010 x
06:5219
06:54EA
06:56was 1 x
06:5810 x 7
07:00we need to multiply
07:02this only
07:041 upon
07:068.5 x
07:081.6
07:101.5
07:12upon
07:1413.6
07:1613.6 x
07:1810 x
07:207
07:2210 x
07:2426
07:2613.6
07:280.110
07:3010 x
07:322
07:341.1
07:3610 x
07:383
07:40meter
07:42per second
07:44it could be written as
07:461.1 x
07:48millimetre
07:50millimetre
07:52per second
07:54second part
07:56second part
07:58A part
08:00average speed
08:02average drift speed
08:04B part
08:06they are asking us
08:08to compare the drift speed obtained
08:10above with
08:12of copper atoms at
08:14ordinary temperatures
08:20first
08:22in part B
08:24we have to find thermal speed
08:26and
08:28one formula from the chapter
08:30kinetic theory of gases
08:32in eleventh
08:34and
08:36there we have to
08:38equate half mv square
08:40with kbt
08:42we got
08:44a formula
08:46that was
08:48thermal speed
08:50rms speed
08:52of electron
08:54having the kinetic energy
08:56half mv square
08:58half mv square
09:00is equal to
09:023 by 2 kbt
09:04where t was
09:06temperature
09:08mass v was
09:10speed
09:12same speed
09:14kb was
09:16Boltzmann constant
09:18you are right
09:20then we
09:22got v
09:24as kbt
09:26by m
09:28from this
09:30thermal speed of copper
09:32approximately
09:34this is not talked
09:36it is roughly
09:38taken
09:40room temperature
09:42and lastly
09:44calculation of m
09:463 kbt
09:481.38 into
09:5010 raise to minus 23
09:52300
09:54on
09:56raise to minus
09:583 on 6
10:00into 10 raise to
10:02it will come over here
10:04it will be directly cancelled
10:06this factor i am directly cancelling
10:08right
10:10then there after
10:12is
10:146 into
10:161.38
10:18into
10:20300 on
10:223.5
10:2410 raise to minus 3
10:26it should come
10:28342
10:30it should not come
10:32that is
10:34100
10:36354
10:38is equal to
10:40in
10:42comparison
10:44comparison
10:461.1 into
10:4810 raise to minus 3
10:50meter per second
10:52ratio required
10:54ratio would be
10:56of
10:58upon
11:00PRMS
11:02of per atom
11:04and
11:06342.57
11:08and after calculation
11:10right now we are
11:12into second part of this question
11:14and this is
11:16B part
11:18in B part it is first part
11:20in B part it is first part
11:22compare the drift speed
11:24obtained above with
11:26thermal speeds of copper atoms
11:28at ordinary temperature
11:30which we have found
11:32speed of propagation
11:34of electric field along the
11:36conductor which
11:38causes the drift motion
11:40B
11:42second part compare
11:44the speed of propagation
11:46electric field
11:48along the conductor
11:50electric field and this is very
11:52important very interesting point
11:54electric field propagates
11:56along a conductor
11:58with the speed of
12:00speed of EM waves
12:02EMW that is
12:04that means we are to compare
12:06required ratio
12:08would be Vdc
12:10that is equal to
12:12already
12:14we had this
12:161.1
12:18into
12:2010 raise to minus 3 upon
12:22upon 3 into
12:2410 raise to 8
12:26just final answer
12:28would be
12:30again approximately
12:3210 raise to 11
12:34final answer
12:36now for
12:38example 3.2
12:40this is appearing
12:42completely theoretical
12:44but still let's see
12:46there are 5 parts in that
12:48A part in example
12:503.1 that means previous question
12:52the electron drift speed
12:54is estimated to be only
12:56a few mm per second for current
12:58in the range of a few
13:00amperes how then
13:02is current established
13:04almost the instant
13:06a circuit is closed
13:08that means
13:10how current is generated so fast
13:12again
13:14for this
13:16factor is electric field
13:18the deciding factor is electric
13:20field and in previous
13:22question B part
13:24B2 second
13:26part we had this
13:28electric field in numerical
13:30format now
13:32same electric field we are
13:34to discuss for
13:36this first part
13:38so for this part
13:40for current generation
13:42they don't have to wait for electrons
13:44then second
13:46part the electron
13:48drift arises due to the
13:50force experienced by electrons
13:52in the electric field inside the conductor
13:54but force should
13:56cause acceleration
13:58why then do the electrons
14:00acquire a steady average drift
14:02this is basic fundamental
14:04for
14:06acquisition of drift velocity
14:08let's come to
14:10that means we need to explain
14:12the reason behind
14:14this average
14:16despite uniformly
14:18accelerated motion
14:20does accelerate
14:22drift speed collides
14:24positive ion
14:26and this
14:28collision only cause
14:30obstruction that is resistance
14:32inside the conductor
14:34it loses its drift speed
14:36after collision
14:38but starts to accelerate
14:40increasing its
14:42drift speed therefore
14:44electrons acquire
14:48and now
14:50third part
14:52if the electron drift
14:54speed is so small
14:56and the electrons charge is small
14:58how can we still
15:00obtain large amount
15:02of current in a conductor
15:04this is again very thoughtful
15:06question because the electron
15:08number density is very high
15:10in the order of 10 raise
15:12to 29 per cubic
15:14meter the reason
15:16further I am to
15:18give one reference
15:20I is equal to
15:22N E A
15:24V D that means
15:26current is depending
15:28upon N also
15:30and N
15:32is this much I
15:34then come to
15:36D part when electron
15:38drift in a metal from
15:40lower to higher potential
15:42does it mean that
15:44all the free electrons of the metal
15:46are moving in the same
15:48direction yes
15:50that is obvious
15:52it has to this was
15:54D part
15:56the drift velocity
15:58is superposed
16:00then come to E part
16:02are the parts of
16:04electrons straight lines
16:06between successive collisions
16:08positive ions of the metal
16:10in the first
16:12absence of electric field presence
16:14of electric field that means in both
16:16the situation it is
16:18it is been asked
16:20whether the movement
16:22of electrons path
16:24are a straight line
16:26or not
16:28in both the situation absence
16:30of electric field may be and
16:32presence may be E part
16:34electric field
16:36the paths are straight lines
16:38and hence this
16:40question number 3.2
16:42is over
16:443.3 tan electric
16:46toaster uses
16:48nichrome for its heating
16:50element when a negligibly
16:52small current passes
16:54through it its resistance
16:56at room temperature
16:58above that means
17:00R T
17:02equals 75
17:040.3 ohms
17:06and where
17:08T is
17:1027 degree Celsius
17:12and this is room temperature
17:14when the toaster is connected to a
17:16230 volt supply
17:18the current settles
17:20after a few seconds
17:22to a steady value
17:24of 2.68
17:262.68
17:28ampere then I is
17:302.68
17:32ampere what is the
17:34steady temperature of the nichrome
17:36element
17:38temperature
17:40the temperature coefficient of resistance
17:42of nichrome averaged
17:44over the temperature range
17:46involved is
17:48that means they are
17:50talking about alpha
17:52alpha is given
17:54temperature coefficient
17:56alpha
17:581.70
18:00into
18:0210 raise to minus 4
18:04degree per
18:06Celsius so this
18:08was alpha and one
18:10more thing what is the steady
18:12temperature of the nichrome
18:14element any final
18:16temperature then
18:18T1 then
18:20T2 is to be
18:22found but
18:24let's find batka
18:26resistance also from
18:28given data R2
18:30could be found R2
18:32equals 230
18:34volt
18:36on 2.68
18:38ampere
18:40that is
18:4285.82
18:44now for
18:46T2 how do we find
18:48now considering
18:50for your again
18:52I am to remind
18:54RT is equal to
18:56R0 1 plus
18:58alpha T
19:00but here
19:02alpha at
19:040 degree Celsius
19:06was not given
19:08alternative formula
19:10R2 is equal to R0
19:121 plus
19:14alpha delta T
19:16has to be used
19:18this is
19:20not to be used then
19:22coming back this
19:24equation R2
19:26is equal to R0
19:281 plus alpha
19:30delta T that means
19:32T2 minus
19:34T1
19:36T2 minus T1
19:38R2 minus R1
19:40upon
19:42R1 alpha
19:44R2 is
19:4685.8
19:48minus R1
19:5075.3
19:5275.3
19:54into
19:56alpha
19:58was 1.70
20:00into 10 raise to minus 4
20:02it implies that
20:04T2 minus T1
20:06is
20:08820 degree Celsius
20:10and already
20:12we had T1
20:1427 degree Celsius
20:16here you can notice
20:18and we are to find T2
20:20820
20:24plus
20:2627
20:28847 degree
20:30Celsius
20:32now for example 3.4
20:34again
20:36it is the question
20:38is related to
20:40the relation between resistance
20:42and temperature only
20:44let's see
20:46what is given the resistance
20:48of platinum wire
20:50of a
20:52platinum
20:54resistance thermometer
20:56at the ice point
20:58at ice point
21:00resistance is
21:025 ohm
21:04R0
21:065 ohm and
21:08at steam point
21:10that means R100
21:12equals
21:145.23
21:16when the thermometer is
21:18inserted in a hot bath
21:20the resistance of the
21:22platinum wire is
21:245.795
21:26RT let's assume
21:28795
21:30ohm calculate the
21:32temperature of the bath
21:34this time temperature
21:36is asked now again
21:38we would think for
21:40two relations
21:42first is
21:44RT is equal to
21:46R0 1 plus
21:48alpha
21:50T second
21:52is R2 is equal
21:54to R0
21:561 plus alpha delta
21:58T now let's see
22:00what is to be taken
22:02going to be noted no
22:04alpha is given
22:06applying this is equation
22:08number one second situation
22:10RT
22:12RT is equal to
22:14R0
22:161 plus alpha
22:20we have
22:22two equations
22:24finally
22:26equation number one and
22:28two in both the equations
22:30we have to
22:32get the value of T
22:34therefore T is
22:36equal to
22:38100 into
22:40what is RT
22:42RT was
22:445.795
22:46and what
22:48was R0
22:50R0 is
22:525 what is
22:54R100 5.23
22:56only
22:58calculation left
23:005.795
23:02minus
23:045
23:06that is
23:08795
23:10upon
23:120.23
23:14get me the final
23:16value
23:183.45 multiplied
23:20by
23:22345
23:24T is
23:26345
23:280.65 degree
23:30Celsius hence
23:32this is done see you
23:34in next video bye for now
23:36take care
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